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>
using namespace std;
const int N = 1000010;
int n;
char s[N];
int cur[N], sum = 0, dp[N];
vector<int> v;
int main() {
  scanf("%s", s);
  n = strlen(s);
  int cnt = 0;
  for (int i = 0; i < n; i++) {
    if (s[i] == '1') {
      if (i != 0) v.push_back(cnt);
      cnt = 0;
    } else
      cnt++;
  }
  v.push_back(cnt);
  dp[(int)v.size() - 1] = v.back() + 1;
  if ((int)v.size() == 1) {
    if (s[0] == '0') {
      cout << n << endl;
    } else
      cout << n + 1 << endl;
    return 0;
  }
  for (int i = (int)v.size() - 2; i >= 0; i--) {
    for (int j = 0; j <= v[i]; j++) {
      sum -= cur[j];
      if (sum < 0) sum += 1000000007;
      cur[j] = dp[i + 1];
      sum += cur[j];
      if (sum >= 1000000007) sum -= 1000000007;
    }
    if (i == 0 && s[i] == '0') {
      dp[i] = (long long)(v[i] + 1) * dp[i + 1] % 1000000007;
      break;
    }
    dp[i] = sum;
    if (i < (int)v.size() - 1) {
      dp[i] += dp[(int)v.size() - 1];
      if (dp[i] >= 1000000007) dp[i] -= 1000000007;
    }
    if (i == 0 && s[i] == '0') {
      dp[i]--;
      if (dp[i] < 0) dp[i] += 1000000007;
    }
  }
  cout << dp[0] << endl;
  return 0;
}
