Problem: There is a string S of length N consisting of characters `0` and `1`. You will perform the following operation for each i = 1, 2, ..., m:

* Arbitrarily permute the characters within the substring of S starting at the l_i-th character from the left and extending through the r_i-th character.



Here, the sequence l_i is non-decreasing.

How many values are possible for S after the M operations, modulo 1000000007(= 10^9+7)?

Constraints

* 2≦N≦3000
* 1≦M≦3000
* S consists of characters `0` and `1`.
* The length of S equals N.
* 1≦l_i < r_i≦N
* l_i ≦ l_{i+1}

Input

The input is given from Standard Input in the following format:


N M
S
l_1 r_1
:
l_M r_M


Output

Print the number of the possible values for S after the M operations, modulo 1000000007.

Examples

Input

5 2
01001
2 4
3 5


Output

6


Input

9 3
110111110
1 4
4 6
6 9


Output

26


Input

11 6
00101000110
2 4
2 3
4 7
5 6
6 10
10 11


Output

143

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

using namespace std;
const int MOD = 1e9 + 7;

void act(int& a, int b) {
    a += b;
    if (a >= MOD)
        a -= MOD;
}

int main() {
    int n, m; cin >> n >> m;

    string s; cin >> s;
    vector<int> ones(n), zero(n);
    for (int i = 0; i < n; ++i) {
        if (s[i] == '1') ones[i] = 1;
        else zero[i] = 1;
        if (i > 0) {
            ones[i] += ones[i - 1];
            zero[i] += zero[i - 1];
        }
    }

    vector<int> best(n);
    for (int i = 0; i < n; ++i)
        best[i] = i;

    for (int i = 0; i < m; ++i) {
        int l, r; cin >> l >> r; --l; --r;
        best[l] = max(best[l], r);
    }
    for (int i = 0; i + 1 < n; ++i) {
        best[i + 1] = max(best[i + 1], best[i]);
    }
    vector<vector<int>> dp(n + 1, vector<int>(n + 1, 0));
    dp[0][0] = 1;
    for (int i = 0; i <= n; ++i) {
        for (int j = 0; i + j <= n; ++j) {
            int at = i + j;
            bool can_zero = zero[best[at]] > i;
            bool can_ones = ones[best[at]] > j;
            if (can_zero) act(dp[i + 1][j], dp[i][j]);
            if (can_ones) act(dp[i][j + 1], dp[i][j]);
        }
    }
    
    int ans = 0;
    for (int i = 0, j = n; i <= n; ++i, --j)
        act(ans, dp[i][j]);
    cout << ans << endl;

    return 0;
}