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 <cstdio>
#include <queue>
#define MAXN 3005
#define mod 1000000007
using namespace std;

int n, m, dp[MAXN][MAXN], cnt0[MAXN];
int qr[MAXN], qrMax[MAXN];
char s[MAXN];
bool uz[MAXN][MAXN];

int getCnt0Pos(int pos) {
    return cnt0[qrMax[pos]] - cnt0[pos - 1];
}

int computeDp(int l, int c) {
    if(l == n) return 1;
    if(uz[l][c]) return dp[l][c];

    int r = qrMax[l];

    int nextC = c + cnt0[qrMax[l + 1]] - cnt0[qrMax[l]];
    
    dp[l][c] = 0;
    if(c < r - l + 1)
        dp[l][c] += computeDp(l + 1, nextC);
    if(c > 0)
        dp[l][c] += computeDp(l + 1, nextC - 1);

    while(dp[l][c] >= mod) dp[l][c] -= mod;
    uz[l][c] = 1;

    return dp[l][c];
}

int main() {

    scanf("%d %d\n", &n, &m);
    scanf("%s", s + 1);

    for(int i = 1; i <= n; ++i)
        cnt0[i] = cnt0[i - 1] + (s[i] == '0');

    for(int i = 1; i <= n; ++i)
        qr[i] = i;

    for(int i = 1; i <= m; ++i) {
        int l, r;
        scanf("%d %d", &l, &r);
        qr[l] = max(qr[l], r);
    }

    for(int i = 1; i <= n; ++i)
        qrMax[i] = max(qrMax[i - 1], qr[i]);

    printf("%d\n", computeDp(1, getCnt0Pos(1)));

    return 0;
}