#include <bits/stdc++.h>

#include <atcoder/modint>

#include <algorithm>
#include <cassert>
#include <vector>

template <class T>
bool chmax(T &a, const T &b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

template <class T>
bool chmin(T &a, const T &b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}

template <class T>
T div_floor(T a, T b) {
    if (b < 0) a *= -1, b *= -1;
    return a >= 0 ? a / b : (a + 1) / b - 1;
}

template <class T>
T div_ceil(T a, T b) {
    if (b < 0) a *= -1, b *= -1;
    return a > 0 ? (a - 1) / b + 1 : a / b;
}

template <typename T>
struct CoordComp {
    std::vector<T> v;
    bool sorted;

    CoordComp() : sorted(false) {}

    int size() { return v.size(); }

    void add(T x) { v.push_back(x); }

    void build() {
        std::sort(v.begin(), v.end());

        v.erase(std::unique(v.begin(), v.end()), v.end());
        sorted = true;
    }

    int get_idx(T x) {
        assert(sorted);
        return lower_bound(v.begin(), v.end(), x) - v.begin();
    }

    T &operator[](int i) { return v[i]; }
};


#define For(i, a, b) for (int i = (int)(a); (i) < (int)(b); ++(i))
#define rFor(i, a, b) for (int i = (int)(a)-1; (i) >= (int)(b); --(i))
#define rep(i, n) For(i, 0, n)
#define rrep(i, n) rFor(i, n, 0)
#define fi first
#define se second

using namespace std;

using lint = long long;
using pii = pair<int, int>;
using pll = pair<lint, lint>;

using namespace atcoder;
using mint = modint1000000007;

int main() {
    int n;
    string s;
    cin >> n >> s;

    string t = "yukicoder";
    vector<vector<mint>> dp(2, vector<mint>(10, 0));
    dp[0][0] = 1;
    rep(i, n) {
        vector<vector<mint>> nxt(2, vector<mint>(10, 0));
        rep(j, 2) rep(k, 10) if (dp[j][k] != 0) {
            nxt[j][k] += dp[j][k];
            if (k + 1 < 10 && s[i] == '?' && j == 0) nxt[1][k + 1] += dp[j][k];
            if (k + 1 < 10 && s[i] == t[k]) nxt[j][k + 1] += dp[j][k];
        }
        swap(dp, nxt);
    }
    mint ans = dp[0][9] + dp[1][9];
    printf("%u\n", ans.val());
}