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.
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:1000000000")
#pragma GCC optimize("Ofast,no-stack-protector")
using namespace std;
const int maxn = (int)1e6 + 10, mod = (int)1e9 + 7;
int add(int x, int y) {
  x += y;
  if (x >= mod) {
    x -= mod;
  }
  return x;
}
int sub(int x, int y) {
  x -= y;
  if (x < 0) {
    x += mod;
  }
  return x;
}
char s[maxn];
int dp[maxn][2];
int sum[maxn];
int cnt[maxn];
int getsum(int l, int r) { return sub(sum[l], sum[r + 1]); }
int main() {
  scanf("%s", s);
  int n = 0;
  while (s[n]) {
    n++;
  }
  int lst[2] = {n, n};
  for (int i = n - 1; i >= 0; i--) {
    if (s[i] == '1') {
      cnt[i] = 0;
    } else {
      cnt[i] = cnt[i + 1] + 1;
    }
  }
  vector<int> q;
  int numb = 0;
  for (int i = n - 1; i >= 0; i--) {
    if (s[i] == '1') {
      break;
    }
    numb++;
  }
  for (int i = n - 1; i >= 0; i--) {
    if (s[i] == '0') {
      dp[i][0] = add(dp[i + 1][0], dp[i + 1][1]);
      if (lst[1] != n) {
        dp[i][1] = dp[lst[1]][1];
      } else {
        dp[i][0] = add(dp[i][0], 1);
      }
      if (i < n - numb) {
        while (!q.empty() && cnt[i] > cnt[q.back()]) {
          q.pop_back();
        }
        q.push_back(i);
      }
    } else {
      dp[i][1] = 1;
      dp[i][1] = add(dp[i][1], numb);
      dp[i][1] = add(dp[i][1], dp[i + 1][1]);
      int prev = 0;
      for (int j = (int)q.size() - 1; j >= 0; j--) {
        int pos = q[j];
        dp[i][1] = add(dp[i][1], dp[pos + prev][0]);
        prev = cnt[pos];
      }
    }
    lst[s[i] - '0'] = i;
    sum[i] = sum[i + 1] + dp[i][0];
  }
  cout << add(dp[0][0], dp[0][1]) << '\n';
}
