#include <bits/stdc++.h>
using namespace std;
const int N = 1e2 + 5;
long long x, y;
int xb[N], yb[N], lenx, leny;
void cmp() {
  for (int i = 1; i <= leny - lenx + 1; i++) {
    bool vis = 1;
    for (int j = 1; j <= lenx; j++) {
      if (xb[j] != yb[i + j - 1]) {
        vis = 0;
      }
    }
    for (int j = 1; j < i; j++) {
      if (yb[j] == 0) {
        vis = 0;
      }
    }
    for (int j = i + lenx; j <= leny; j++) {
      if (yb[j] == 0) {
        vis = 0;
      }
    }
    if (xb[1] == 0 && i - 1 == 0 && leny - (i + lenx - 1) > 0) {
      vis = 0;
    }
    if (vis) {
      printf("YES");
      exit(0);
    }
  }
}
int main() {
  scanf("%lld%lld", &x, &y);
  while (x) {
    xb[++lenx] = x % 2;
    x /= 2;
  }
  while (y) {
    yb[++leny] = y % 2;
    y /= 2;
  }
  cmp();
  reverse(yb + 1, yb + leny + 1);
  cmp();
  reverse(xb + 1, xb + lenx + 1);
  while (!xb[lenx]) {
    lenx--;
  }
  cmp();
  reverse(yb + 1, yb + leny + 1);
  cmp();
  printf("NO");
  return 0;
}

### Prompt

In CPP, your task is to solve the following problem:

You are given two positive integers x and y. You can perform the following operation with x: write it in its binary form without leading zeros, add 0 or 1 to the right of it, reverse the binary form and turn it into a decimal number which is assigned as the new value of x.

For example: 

  * 34 can be turned into 81 via one operation: the binary form of 34 is 100010, if you add 1, reverse it and remove leading zeros, you will get 1010001, which is the binary form of 81. 
  * 34 can be turned into 17 via one operation: the binary form of 34 is 100010, if you add 0, reverse it and remove leading zeros, you will get 10001, which is the binary form of 17. 
  * 81 can be turned into 69 via one operation: the binary form of 81 is 1010001, if you add 0, reverse it and remove leading zeros, you will get 1000101, which is the binary form of 69. 
  * 34 can be turned into 69 via two operations: first you turn 34 into 81 and then 81 into 69. 



Your task is to find out whether x can be turned into y after a certain number of operations (possibly zero).

Input

The only line of the input contains two integers x and y (1 ≤ x, y ≤ 10^{18}).

Output

Print YES if you can make x equal to y and NO if you can't.

Examples

Input


3 3


Output


YES


Input


7 4


Output


NO


Input


2 8


Output


NO


Input


34 69


Output


YES


Input


8935891487501725 71487131900013807


Output


YES

Note

In the first example, you don't even need to do anything.

The fourth example is described in the statement.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1e2 + 5;
long long x, y;
int xb[N], yb[N], lenx, leny;
void cmp() {
  for (int i = 1; i <= leny - lenx + 1; i++) {
    bool vis = 1;
    for (int j = 1; j <= lenx; j++) {
      if (xb[j] != yb[i + j - 1]) {
        vis = 0;
      }
    }
    for (int j = 1; j < i; j++) {
      if (yb[j] == 0) {
        vis = 0;
      }
    }
    for (int j = i + lenx; j <= leny; j++) {
      if (yb[j] == 0) {
        vis = 0;
      }
    }
    if (xb[1] == 0 && i - 1 == 0 && leny - (i + lenx - 1) > 0) {
      vis = 0;
    }
    if (vis) {
      printf("YES");
      exit(0);
    }
  }
}
int main() {
  scanf("%lld%lld", &x, &y);
  while (x) {
    xb[++lenx] = x % 2;
    x /= 2;
  }
  while (y) {
    yb[++leny] = y % 2;
    y /= 2;
  }
  cmp();
  reverse(yb + 1, yb + leny + 1);
  cmp();
  reverse(xb + 1, xb + lenx + 1);
  while (!xb[lenx]) {
    lenx--;
  }
  cmp();
  reverse(yb + 1, yb + leny + 1);
  cmp();
  printf("NO");
  return 0;
}

```