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;
inline int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  return x * f;
}
inline void print(int x) {
  if (x < 0) x = -x, putchar('-');
  if (x >= 10) print(x / 10);
  putchar(x % 10 + 48);
}
int n;
const int mod = 1e9 + 7;
char s[1000010];
int nxt[1000010];
int dp[1000010];
int len[1000010];
int nxt1[1000010], nxtd[1000010], lst[1000010];
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  int t = n + 1;
  for (int i = n; i >= 1; --i) {
    nxt1[i] = t;
    if (s[i] == '1') t = i;
  }
  for (int i = 1; i <= n; ++i) {
    lst[i] = n + 1;
    if (s[i] == '0') len[i] = len[i - 1] + 1;
  }
  for (int i = n; i >= 1; --i) {
    if (s[i] == '0') {
      nxtd[i] = lst[len[i] + 1];
      lst[len[i]] = i;
    } else
      nxtd[i] = lst[1];
  }
  int st = 0;
  for (int i = 1; i <= n; ++i) {
    if (s[i] == '1') {
      st = i;
      break;
    }
  }
  if (st == 0) {
    printf("%d\n", n);
    return 0;
  }
  dp[st] = st;
  for (int i = st; i < n; ++i) {
    if (nxt1[i] != n + 1) {
      dp[nxt1[i]] = (dp[nxt1[i]] + dp[i]) % mod;
    }
    if (nxtd[i] != n + 1) {
      dp[nxtd[i]] = (dp[nxtd[i]] + dp[i]) % mod;
    }
  }
  int ans = 0;
  for (int i = st; i <= n; ++i) {
    if (len[i] <= len[n]) ans = (ans + dp[i]) % mod;
  }
  printf("%d\n", ans);
  return 0;
}