Problem: JATC loves Banh-mi (a Vietnamese food). His affection for Banh-mi is so much that he always has it for breakfast. This morning, as usual, he buys a Banh-mi and decides to enjoy it in a special way.

First, he splits the Banh-mi into n parts, places them on a row and numbers them from 1 through n. For each part i, he defines the deliciousness of the part as x_i ∈ \{0, 1\}. JATC's going to eat those parts one by one. At each step, he chooses arbitrary remaining part and eats it. Suppose that part is the i-th part then his enjoyment of the Banh-mi will increase by x_i and the deliciousness of all the remaining parts will also increase by x_i. The initial enjoyment of JATC is equal to 0.

For example, suppose the deliciousness of 3 parts are [0, 1, 0]. If JATC eats the second part then his enjoyment will become 1 and the deliciousness of remaining parts will become [1, \\_, 1]. Next, if he eats the first part then his enjoyment will become 2 and the remaining parts will become [\\_, \\_, 2]. After eating the last part, JATC's enjoyment will become 4.

However, JATC doesn't want to eat all the parts but to save some for later. He gives you q queries, each of them consisting of two integers l_i and r_i. For each query, you have to let him know what is the maximum enjoyment he can get if he eats all the parts with indices in the range [l_i, r_i] in some order.

All the queries are independent of each other. Since the answer to the query could be very large, print it modulo 10^9+7.

Input

The first line contains two integers n and q (1 ≤ n, q ≤ 100 000).

The second line contains a string of n characters, each character is either '0' or '1'. The i-th character defines the deliciousness of the i-th part.

Each of the following q lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of the corresponding query.

Output

Print q lines, where i-th of them contains a single integer — the answer to the i-th query modulo 10^9 + 7.

Examples

Input

4 2
1011
1 4
3 4


Output

14
3


Input

3 2
111
1 2
3 3


Output

3
1

Note

In the first example: 

  * For query 1: One of the best ways for JATC to eats those parts is in this order: 1, 4, 3, 2. 
  * For query 2: Both 3, 4 and 4, 3 ordering give the same answer. 



In the second example, any order of eating parts leads to the same answer.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long sz = 2e5 + 5;
const long long MD = 1e9 + 7;
pair<long long, long long> st[2 * sz];
long long arr[sz];
long long n, q;
string s;
pair<long long, long long> ADD(pair<long long, long long> &A,
                               pair<long long, long long> &B) {
  return {A.first + B.first, A.second + B.second};
}
void build() {
  for (long long i = 0; i < n; ++i) {
    arr[i] = s[i] == '1';
  }
  for (long long i = 0; i < n; ++i) {
    st[i + n] = {arr[i] == 0, arr[i] == 1};
  }
  for (long long i = n - 1; i >= 0; --i) {
    st[i] = ADD(st[i << 1], st[i << 1 | 1]);
  }
}
pair<long long, long long> query(long long L, long long R) {
  --L;
  pair<long long, long long> tot = {0LL, 0LL};
  for (L += n, R += n; L < R; L >>= 1, R >>= 1) {
    if (L & 1) tot = ADD(tot, st[L++]);
    if (R & 1) tot = ADD(tot, st[--R]);
  }
  return tot;
}
long long powMod(long long a, long long p) {
  long long res = 1;
  for (; p; p >>= 1, a = ((a) * (a)) % MD) {
    if (p & 1) res *= a, res %= MD;
  }
  return res;
}
long long MUL(long long a, long long b) {
  a %= MD;
  b %= MD;
  return (a * b) % MD;
}
int main() {
  scanf("%lld", &n), scanf("%lld", &q);
  cin >> s;
  build();
  while (q--) {
    long long L, R;
    scanf("%lld", &L), scanf("%lld", &R);
    pair<long long, long long> res = query(L, R);
    long long le = (powMod(2LL, res.second) + MD - 1) % MD;
    long long ri = (powMod(2LL, res.first) + MD - 1) % MD;
    printf("%lld\n", MUL(le, 1LL + ri));
  }
}