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;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const int mod = 1e9 + 7;
const int N = 1e6 + 10;
bool f1, f2;
int a[N], n, m, b[N << 2], sum[N], pos[N], dp[N];
char s[N];
int qry(int l, int r, int rt, int x, int y) {
  if (x <= l && r <= y) return b[rt];
  int mid = l + r >> 1;
  int ans = 0;
  if (x <= mid) ans = max(ans, qry(l, mid, rt << 1, x, y));
  if (y > mid) ans = max(ans, qry(mid + 1, r, rt << 1 | 1, x, y));
  return ans;
}
void modify(int l, int r, int rt, int x, int y) {
  if (l == r) {
    b[rt] = max(b[rt], y);
    return;
  }
  int mid = l + r >> 1;
  if (x <= mid)
    modify(l, mid, rt << 1, x, y);
  else
    modify(mid + 1, r, rt << 1 | 1, x, y);
  b[rt] = max(b[rt << 1], b[rt << 1 | 1]);
}
inline int lowbit(int x) { return x & (-x); }
int qry(int x) {
  int ret = 0;
  while (x < N) {
    ret = max(ret, b[x]);
    x += x & (-x);
  }
  return ret;
}
void modify(int x, int y) {
  while (x) {
    b[x] = max(b[x], y);
    x -= x & (-x);
  }
}
int main() {
  scanf("%s", s + 1);
  m = strlen(s + 1);
  for (int i = 1; i <= m; i++)
    if (s[i] == '1') pos[++n] = i;
  if (n == 0) {
    cout << m << endl;
    return 0;
  }
  for (int i = 1; i <= n; i++) a[i] = pos[i] - pos[i - 1] - 1;
  a[n + 1] = m - pos[n];
  n++;
  int res = 1ll * (a[1] + 1) * (a[n] + 1) % mod;
  n -= 2;
  for (int i = 1; i <= n; i++) a[i] = a[i + 1];
  dp[0] = 1;
  sum[0] = 1;
  for (int i = 1; i <= n; i++) {
    dp[i] = dp[i - 1];
    for (int j = 1; j <= a[i]; j++) {
      int x = qry(j);
      (dp[i] += sum[i - 1]) %= mod;
      if (x > 0) dp[i] -= sum[x - 1];
      if (dp[i] < 0) dp[i] += mod;
      dp[i] %= mod;
    }
    sum[i] = sum[i - 1];
    (sum[i] += dp[i]) %= mod;
    modify(a[i], i);
  }
  printf("%lld\n", 1ll * sum[n] * res % mod);
  return 0;
}
