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
#include <bits/stdc++.h>
using namespace std;
const int N = 3010;
const int MOD = 1e9 + 7;
int n, m;
int li[N], ri[N];
int C[N][N], f[N][N];
string st;
void ADD(int &t, int d) {t += d; if (t >= MOD) t -= MOD;}
int main()
{
    cin >> n >> m;
    cin >> st;
    for (int i = 0; i <= n; ++ i)
    {
        C[i][0] = C[i][i] = 1;
        for (int j = 1; j < i; ++ j)
            C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % MOD;
    }
    for (int i = 1; i <= m; ++ i)
    {
        cin >> li[i] >> ri[i];
        if (ri[i] < ri[i - 1]) i --, m --;
    }
    li[m + 1] = ri[m + 1] = n + 1;
    f[0][0] = 1;
    for (int i = 1; i <= m; ++ i)
    {
        int c = 0;
        for (int j = max(ri[i - 1] + 1, li[i]); j <= ri[i]; ++ j)
            if (st[j - 1] == '1') c ++;
        for (int j = 0; j <= max(ri[i - 1] - li[i] + 1, 0); ++ j)
            for (int k = 0; k <= min(j + c, min(li[i + 1] - 1, ri[i]) - li[i] + 1); ++ k)
                if (j + c - k <= max(ri[i] - li[i + 1] + 1, 0))
                    ADD(f[i][j + c - k], 1ll * f[i - 1][j] * C[min(li[i + 1] - 1, ri[i]) - li[i] + 1][k] % MOD);
    }
    cout << f[m][0] << endl;
}
