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 MOD = (int)1e9 + 7;
void sadd(int &a, int b) {
  a += b;
  if (a >= MOD) a -= MOD;
}
int add(int a, int b) {
  sadd(a, b);
  return a;
}
void ssub(int &a, int b) {
  a -= b;
  if (a < 0) a += MOD;
}
int sub(int a, int b) {
  ssub(a, b);
  return a;
}
int mul(int a, int b) { return (a * 1LL * b) % MOD; }
const int N = (int)1e6 + 100;
char s[N];
int clen[N];
int slen[N];
int dp[N];
int main(int, char **) {
  scanf("%s", s);
  int n = strlen(s);
  int len = 0;
  for (int i = 0; i < n; i++) {
    if (s[i] == '1') {
      len = 0;
    } else {
      clen[i] = len++;
    }
  }
  int ans = 0;
  int suf0 = 0;
  while (suf0 < n && s[n - 1 - suf0] == '0') suf0++;
  if (suf0 == n) {
    printf("%d\n", n);
    return 0;
  }
  int pref0 = 0;
  while (s[pref0] == '0') pref0++;
  int cur = pref0 + 1;
  int cur0 = 0;
  for (int i = pref0; i < n; i++) {
    if (s[i] == '1') {
      sadd(ans, cur);
      sadd(cur0, cur);
    } else {
      int cc = clen[i];
      int cx = sub(cur0, slen[cc]);
      if (cc < suf0) sadd(ans, cx);
      slen[cc] = cur0;
      sadd(cur, cx);
    }
  }
  printf("%d\n", ans);
  return 0;
}
