Problem: Koa the Koala has a binary string s of length n. Koa can perform no more than n-1 (possibly zero) operations of the following form:

In one operation Koa selects positions i and i+1 for some i with 1 ≤ i < |s| and sets s_i to max(s_i, s_{i+1}). Then Koa deletes position i+1 from s (after the removal, the remaining parts are concatenated).

Note that after every operation the length of s decreases by 1.

How many different binary strings can Koa obtain by doing no more than n-1 (possibly zero) operations modulo 10^9+7 (1000000007)?

Input

The only line of input contains binary string s (1 ≤ |s| ≤ 10^6). For all i (1 ≤ i ≤ |s|) s_i = 0 or s_i = 1.

Output

On a single line print the answer to the problem modulo 10^9+7 (1000000007).

Examples

Input


000


Output


3


Input


0101


Output


6


Input


0001111


Output


16


Input


00101100011100


Output


477

Note

In the first sample Koa can obtain binary strings: 0, 00 and 000.

In the second sample Koa can obtain binary strings: 1, 01, 11, 011, 101 and 0101. For example:

  * to obtain 01 from 0101 Koa can operate as follows: 0101 → 0(10)1 → 011 → 0(11) → 01. 
  * to obtain 11 from 0101 Koa can operate as follows: 0101 → (01)01 → 101 → 1(01) → 11. 



Parentheses denote the two positions Koa selected in each operation.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1000000007;
char S[1000005];
int n;
int m;
int A[1000005];
int nextGreater[1000005];
long long dp[1000005];
long long tempSum[1000005];
int main() {
  scanf(" %s", S);
  n = strlen(S);
  int streak = 0;
  vector<int> tempA;
  for (int i = 0; i < n; i++) {
    if (S[i] == '1') {
      tempA.push_back(streak);
      streak = 0;
    } else if (S[i] == '0') {
      streak++;
    }
  }
  tempA.push_back(streak);
  if ((int)tempA.size() == 1) {
    printf("%d\n", n);
    return 0;
  }
  m = (int)tempA.size() - 2;
  for (int i = 1; i <= m; i++) {
    A[i - 1] = tempA[i];
  }
  for (int i = 0; i < m; i++) {
    nextGreater[i] = m;
  }
  stack<pair<int, int> > stack1;
  for (int i = 0; i < m; i++) {
    while (!stack1.empty()) {
      pair<int, int> temp = stack1.top();
      if (A[i] > temp.first) {
        nextGreater[temp.second] = i;
        stack1.pop();
      } else {
        break;
      }
    }
    stack1.push(pair<int, int>(A[i], i));
  }
  for (int i = 0; i < m; i++) {
  }
  long long initWays =
      ((long long)(tempA.front() + 1LL) * (long long)(tempA.back() + 1LL)) %
      MOD;
  dp[m] = initWays;
  memset(tempSum, -1, sizeof(tempSum));
  for (int i = m - 1; i >= 0; i--) {
    dp[i] = ((A[i] + 1) * dp[i + 1] + initWays) % MOD;
    tempSum[i] = 0;
    int curIndx = i;
    while (curIndx < m) {
      int nextIndx = nextGreater[curIndx];
      if (nextIndx == m) {
        break;
      }
      tempSum[i] += (A[nextIndx] - A[curIndx] + MOD) * dp[nextIndx + 1] % MOD;
      if (tempSum[nextIndx] != -1) {
        tempSum[i] += tempSum[nextIndx];
        break;
      }
      curIndx = nextIndx;
    }
    dp[i] = (tempSum[i] + dp[i]) % MOD;
  }
  printf("%lld", dp[0]);
  return 0;
}