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 = 1000005;
const long long mod = 1000000007ll;
int n, num[N], cnt = 0;
char str[N];
void init() {
  int prv = -1;
  for (int i = 0; i < n; i++) {
    if (str[i] == '1') {
      num[++cnt] = i - prv - 1;
      prv = i;
    }
  }
  num[++cnt] = n - prv - 1;
}
long long f[N], sgt_tag[N << 2], sgt_val[N << 2];
void build(int l, int r, int now) {
  int mid = l + r >> 1;
  sgt_tag[now] = -1ll;
  sgt_val[now] = 0ll;
  if (l < r) build(l, mid, now << 1), build(mid + 1, r, now << 1 | 1);
}
void cover(int l, int r, int now, long long val) {
  sgt_tag[now] = val;
  sgt_val[now] = 1ll * val * (r - l + 1) % mod;
}
void pushdown(int l, int r, int now) {
  int mid = l + r >> 1;
  if (sgt_tag[now] >= 0) {
    cover(l, mid, now << 1, sgt_tag[now]);
    cover(mid + 1, r, now << 1 | 1, sgt_tag[now]);
    sgt_tag[now] = -1;
  }
}
void pushup(int now) {
  sgt_val[now] = (sgt_val[now << 1] + sgt_val[now << 1 | 1]) % mod;
}
void change(int left, int right, int l, int r, int now, long long val) {
  int mid = l + r >> 1;
  if (l == left && r == right)
    cover(l, r, now, val);
  else {
    pushdown(l, r, now);
    if (right <= mid)
      change(left, right, l, mid, now << 1, val);
    else if (left > mid)
      change(left, right, mid + 1, r, now << 1 | 1, val);
    else
      change(left, mid, l, mid, now << 1, val),
          change(mid + 1, right, mid + 1, r, now << 1 | 1, val);
    pushup(now);
  }
}
void solve() {
  f[cnt + 1] = 0ll;
  build(0, n + 1, 1);
  long long now = 0ll;
  for (int i = cnt; i >= 1; i--) {
    change(0, num[i], 0, n + 1, 1, f[i + 1]);
    if (i == 1) {
      if (cnt > 1)
        f[i] = f[i + 1] * (num[i] + 1) % mod;
      else
        f[i] = num[i];
    } else
      f[i] = (sgt_val[1] + num[cnt] + 1) % mod;
  }
  printf("%lld\n", f[1]);
}
int main() {
  scanf("%s", str), n = strlen(str);
  init();
  solve();
  return 0;
}
