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>
const int MAXN = 1e6, MOD = 1e9 + 7;
int n, f[MAXN + 5][2], trs[MAXN + 5][2], zcnt, top, stk[MAXN + 5];
char s[MAXN + 5];
std::pair<int, int> zsec[MAXN + 5];
inline void addeq(int& a, const int b) {
  if ((a += b) >= MOD) a -= MOD;
}
int main() {
  scanf("%s", s + 1), n = strlen(s + 1);
  bool allz = true;
  for (int i = 1; i <= n && allz; ++i) allz &= s[i] == '0';
  if (allz) return printf("%d\n", n), 0;
  int L = 0, R = 0;
  for (int i = 1; i <= n && !L; ++i)
    if (s[i] ^ '0') L = i;
  for (int i = n; i && !R; --i)
    if (s[i] ^ '0') R = i;
  for (int i = R, las = R + 1; i >= L; --i) {
    trs[i][1] = las;
    if (s[i] ^ '0') {
      las = i;
      if (s[i + 1] ^ '0') trs[i][0] = trs[i + 1][0];
    }
    if (i == R || s[i + 1] == '0') trs[i][0] = i + 1;
  }
  for (int l = L, r; l <= R; l = r + 1) {
    r = l;
    for (; r <= R && s[r] == '0'; ++r)
      ;
    if (l < r) zsec[++zcnt] = {l, r - l};
  }
  for (int i = zcnt; i; --i) {
    for (; top && zsec[stk[top]].second <= zsec[i].second; --top)
      ;
    trs[zsec[i].first + zsec[i].second - 1][0] =
        top ? zsec[stk[top]].first + zsec[i].second : R + 1;
    stk[++top] = i;
  }
  int ans = 0;
  f[L][1] = 1;
  for (int i = L, s; i <= R; ++i) {
    addeq(s = f[i][0], f[i][1]), addeq(ans, f[i][1]);
    if (trs[i][0] <= R) addeq(f[trs[i][0]][0], s);
    if (trs[i][1] <= R) addeq(f[trs[i][1]][1], s);
  }
  printf("%d\n", int((n - R + 1ll) % MOD * L % MOD * ans % MOD));
  return 0;
}