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;

template<typename T>
void out(T x) { cout << x << endl; exit(0); }
#define watch(x) cout << (#x) << " is " << (x) << endl




typedef long long ll;
const ll mod = 1e9+7;
const int maxn = 3030;



int n, m;
string s;
int a[maxn];
int b[maxn];
ll dp[maxn][maxn];//# ways to fill first i elems, holding j 1's



void add(ll& x, ll y) {
    x %= mod;
    y %= mod;
    x += y;
    x %= mod;
}

int main() {
    ios_base::sync_with_stdio(false); cin.tie(0);  cout.tie(0);
    cin>>n>>m;
    cin>>s;
    for (int i=0; i<n; i++) {
	a[i]=(s[i]-'0');
	b[i]=i+1;
    }

    for (int i=0; i<m; i++) {
	int l,r; cin>>l>>r;
	--l; --r;
	b[l]=max(b[l],r+1);
    }

    for (int i=1; i<n; i++) {
	b[i]=max(b[i],b[i-1]);
    }

    dp[0][0]=1;
    for (int i=0, j=0; i<n; i++) {
	int ones = 0;
	while (j<b[i]) ones+=a[j++];

	for (int x=0; x<=n; x++) {
	    if (dp[i][x] == 0) continue;
	    int cur=x+ones;

	    //put 1 at ith index
	    if (cur>0) {
		add(dp[i+1][cur-1], dp[i][x]);
	    }

	    // put 0 at ith index
	    if (j-i != cur) {
		add(dp[i+1][cur], dp[i][x]);
	    }
	}
    }

    ll ans = dp[n][0];
    ans %= mod;
    ans += mod;
    ans %= mod;
    cout<<ans<<endl;
  

    return 0;
}
