#include <bits/stdc++.h>

using namespace std;

#define rep(i,n) for(int i = 0; i < (int)(n); ++i)
#define rrep(i,n) for(int i = (int)(n) - 1; i >= 0; --i)
#define ALL(a) a.begin(), a.end()
#define Sort(a) sort(a.begin(), a.end())
#define RSort(a) sort(a.rbegin(), a.rend())

typedef long long int ll;
typedef long double ld;
typedef vector<int> vi;
typedef vector<long long> vll;
typedef vector<char> vc;
typedef vector<string> vst;
typedef vector<double> vd;
typedef pair<long long, long long> P;

const long long INF = 0x1fffffffffffffff;
const long long MOD = 998244353;
const long double PI = acos(-1);
 
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
template<class T, class U> inline T vin(T& vec, U n) { vec.resize(n); for(int i = 0; i < (int) n; ++i) cin >> vec[i]; return vec; }
template<class T> inline void vout(T vec, string s = "\n"){ for(auto x : vec) cout << x << s; }
template<class... T> void in(T&... a){ (cin >> ... >> a); }
void out(){ cout << '\n'; }
template<class T, class... Ts> void out(const T& a, const Ts&... b){ cout << a; (cout << ... << (cout << ' ', b)); cout << '\n'; }
template<class T, class U> void inGraph(vector<vector<T>>& G, U n, U m, bool directed = false){ G.resize(n); for(int i = 0; i < m; i++){ int a, b; cin >> a >> b; a--, b--; G[a].push_back(b); if(!directed) G[b].push_back(a); } }

namespace CRT{
    long long extGCD(long long a, long long b, long long &x, long long &y){
        if(b == 0){
            x = 1;
            y = 0;
            return a;
        }
        long long d = extGCD(b, a % b, y, x);
        y -= a / b * x;
        return d;
    }

    pair<long long, long long> ChineseRem(const vector<long long> &b, const vector<long long> &m) {
        long long r = 0, M = 1;
        for(int i = 0; i < (int) b.size(); i++){
            long long p, q;
            long long d = extGCD(M, m[i], p, q);
            if((b[i] - r) % d != 0) return {0, -1};
            long long tmp = (b[i] - r) / d * p % (m[i] / d);
            r += M * tmp;
            M *= m[i] / d;
        }
        r %= M;
        if(r < 0) r += M;
        return {r, M};
    }
}

ll n, m;

void input(){
    in(n, m);
}

void solve(){
    vll b(m), c(m);
    rep(i, m){
        in(b[i], c[i]);
    }
    pair<long long, long long> p = CRT::ChineseRem(c, b);
    if(p.second == -1 || p.first > n){
        cout << "NaN" << endl;
    }else{
        cout << p.first << endl;
    }
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    
    input();
    solve();
}