#pragma gcc target("avx2")
#pragma gcc optimize("o3")
#pragma gcc optimize("unroll-loops")
//#pragma warning(disable : 4996)

//#define ATCODER
#ifdef ATCODER
#include <atcoder/all>
using namespace atcoder;
#endif

#include <limits.h>
#include <math.h>
#include <time.h>

#include <algorithm>
#include <array>
#include <bitset>
#include <cassert>
#include <complex>
#include <cstdio>
#include <cstring>
#include <functional>
#include <iomanip>
#include <iostream>
#include <iterator>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <set>
#include <stack>
#include <string>
#include <type_traits>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>

using namespace std;

#if defined(LOCAL) || defined(_MSC_VER)
void debug_out() { cerr << endl; }

template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
    cerr << " " << to_string(H);
    debug_out(T...);
}

#define DEBUG(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define DEBUG(...) 42
#endif

#ifdef _MSC_VER
#include <intrin.h>

#include "stdafx.h"
#define __builtin_popcount __popcnt
#define __builtin_popcountll __popcnt64
#endif

#define ll long long
#define REP(i, n) for (ll i = 0; i < (n); ++i)
#define REPR(i, n) for (ll i = (n)-1; i >= 0; --i)
#define FOR(i, m, n) for (ll i = (m); i < (n); ++i)
#define VSORT(v) sort(v.begin(), v.end());
#define VREVERSE(v) reverse(v.begin(), v.end())
#define print(x) cout << (x) << '\n'
#define spa << " " <<
#define lb(v, n) lower_bound(v.begin(), v.end(), (n))
#define ub(v, n) upper_bound(v.begin(), v.end(), (n))
#define int long long
//#define double long double
#define all(x) (x).begin(), (x).end()
#define print_space(v) \
    REP(i, v.size()) cout << v[i] << " \n"[i == (int)v.size() - 1]
template <typename T1, typename T2>
inline bool chmin(T1& a, T2 b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template <typename T1, typename T2>
inline bool chmax(T1& a, T2 b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
typedef pair<int, int> pii;
typedef pair<long long, long long> pll;
std::random_device rd;
std::mt19937 mt(rd());
constexpr ll MOD = 1e9 + 7;
constexpr int MAX = 40040;
const double pi = acos(-1);
constexpr long double EPS = 1e-9;
const int dx[4] = { 0, 0, -1, 1 }, dy[4] = { -1, 1, 0, 0 };
using ld = long double;

const int mod = 1000000007;
//const int mod = 998244353;
struct mint {
	ll x; // typedef long long ll;
	mint(ll x = 0) :x((x% mod + mod) % mod) {}
	mint operator-() const { return mint(-x); }
	mint& operator+=(const mint a) {
		if ((x += a.x) >= mod) x -= mod;
		return *this;
	}
	mint& operator-=(const mint a) {
		if ((x += mod - a.x) >= mod) x -= mod;
		return *this;
	}
	mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this; }
	mint operator+(const mint a) const { return mint(*this) += a; }
	mint operator-(const mint a) const { return mint(*this) -= a; }
	mint operator*(const mint a) const { return mint(*this) *= a; }
	mint pow(ll t) const {
		if (!t) return 1;
		mint a = pow(t >> 1);
		a *= a;
		if (t & 1) a *= *this;
		return a;
	}

	// for prime mod
	mint inv() const { return pow(mod - 2); }
	mint& operator/=(const mint a) { return *this *= a.inv(); }
	mint operator/(const mint a) const { return mint(*this) /= a; }
};
istream& operator>>(istream& is, const mint& a) { return is >> a.x; }
ostream& operator<<(ostream& os, const mint& a) { return os << a.x; }

mint fac[MAX], finv[MAX], inv[MAX];
void COMinit() {
	fac[0] = fac[1] = 1;
	finv[0] = finv[1] = 1;
	inv[1] = 1;
	for (int i = 2; i < MAX; i++) {
		fac[i] = fac[i - 1] * i;
		inv[i] = -inv[MOD % i] * (MOD / i);
		finv[i] = finv[i - 1] * inv[i];
	}
}
mint COM(int n, int k) {
	if (n < k) return 0;
	if (n < 0 || k < 0) return 0;
	return fac[n] * (finv[k] * finv[n - k]);
}
mint dp[3030][3030];
void solve() {
	int N, M, K; cin >> N >> M >> K;
	vector<int>L(M), R(M);
	REP(i, M) {
		cin >> L[i] >> R[i];
		L[i]--;
	}
	FOR(i, 1, N + 1)dp[0][i] = 1;
	REP(i, K) {
		REP(j, M) {

			dp[i+1][L[j]+1]+= dp[i][R[j]] - dp[i][L[j]];
			dp[i + 1][R[j]+1] -= dp[i][R[j]] - dp[i][L[j]];
		}
		REP(j, N+1)dp[i + 1][j + 1] += dp[i + 1][j];
		REP(j,N+1)dp[i + 1][j + 1] += dp[i + 1][j];
	}
	print(dp[K][N] - dp[K][N - 1]);
}

signed main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    // REP(_, 100)test();
     //int q;
     //cin >> q;
     //while (q--)
    solve();
}