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 out = 0, fh = 1;
  char jp = getchar();
  while ((jp > '9' || jp < '0') && jp != '-') jp = getchar();
  if (jp == '-') fh = -1, jp = getchar();
  while (jp >= '0' && jp <= '9') out = out * 10 + jp - '0', jp = getchar();
  return out * fh;
}
void print(int x) {
  if (x >= 10) print(x / 10);
  putchar('0' + x % 10);
}
void write(int x, char c) {
  if (x < 0) putchar('-'), x = -x;
  print(x);
  putchar(c);
}
namespace Module {
const int P = 1e9 + 7;
int add(int a, int b) { return a + b >= P ? a + b - P : a + b; }
void inc(int &a, int b) { a = add(a, b); }
int mul(int a, int b) { return 1LL * a * b % P; }
int fpow(int a, int b) {
  int res = 1;
  while (b) {
    if (b & 1) res = mul(res, a);
    a = mul(a, a);
    b >>= 1;
  }
  return res;
}
}  // namespace Module
using namespace Module;
const int N = 1e6 + 10;
int n, dp[N], a[N], len[N], lst, nxt[N], sum[N];
char s[N];
int calc(int x) { return nxt[x] <= n ? dp[nxt[x]] : 0; }
int main() {
  scanf("%s", s), n = strlen(s), lst = -1;
  for (int i = 0; i < n; ++i) {
    sum[i] = sum[i - 1] + (a[i] = s[i] - '0');
    if (a[i] == 0)
      len[i] = i - lst;
    else
      lst = i;
  }
  if (!sum[n - 1]) return printf("%d\n", n), 0;
  for (int i = n - 1; i >= 0; --i) {
    dp[i] = (len[i] <= len[n - 1]) + add(calc(0), calc(len[i] + 1));
    nxt[len[i]] = i;
  }
  int ans = mul(nxt[0] + 1, calc(0));
  cout << ans << '\n';
  return 0;
}