#include <bits/stdc++.h>
#pragma gcc optimize("unroll-loop")
#pragma gcc optimize("Ofast")
using namespace std;
const long long MOD = 1e9 + 7;
const long long N = 1e6 + 5;
long long gcd(long long n, long long m) {
  if (n == 0)
    return m;
  else
    return gcd(m % n, n);
}
long long ans[N], a[N], b[N];
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long n;
  cin >> n;
  string s;
  cin >> s;
  s = '#' + s;
  long long sum = 0;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i] >> b[i];
    long long light_on = 0;
    if (s[i] == '1') light_on = 1;
    for (long long j = 0; j <= 500; j++) {
      if (j < b[i]) {
        if (light_on == 1)
          ans[j] += 1;
        else
          ans[j] += 0;
      } else {
        if ((j - b[i]) % a[i] == 0) {
          ans[j] += light_on ^ 1;
          light_on ^= 1;
        } else {
          ans[j] += light_on;
        }
      }
    }
  }
  for (long long i = 0; i <= 500; i++) {
    sum = max(sum, ans[i]);
  }
  cout << sum << endl;
}

### Prompt

Develop a solution in Cpp to the problem described below:

It is a holiday season, and Koala is decorating his house with cool lights! He owns n lights, all of which flash periodically.

After taking a quick glance at them, Koala realizes that each of his lights can be described with two parameters a_i and b_i. Light with parameters a_i and b_i will toggle (on to off, or off to on) every a_i seconds starting from the b_i-th second. In other words, it will toggle at the moments b_i, b_i + a_i, b_i + 2 ⋅ a_i and so on.

You know for each light whether it's initially on or off and its corresponding parameters a_i and b_i. Koala is wondering what is the maximum number of lights that will ever be on at the same time. So you need to find that out.

<image> Here is a graphic for the first example.

Input

The first line contains a single integer n (1 ≤ n ≤ 100), the number of lights.

The next line contains a string s of n characters. The i-th character is "1", if the i-th lamp is initially on. Otherwise, i-th character is "0".

The i-th of the following n lines contains two integers a_i and b_i (1 ≤ a_i, b_i ≤ 5) — the parameters of the i-th light.

Output

Print a single integer — the maximum number of lights that will ever be on at the same time.

Examples

Input


3
101
3 3
3 2
3 1


Output


2


Input


4
1111
3 4
5 2
3 1
3 2


Output


4


Input


6
011100
5 3
5 5
2 4
3 5
4 2
1 5


Output


6

Note

For first example, the lamps' states are shown in the picture above. The largest number of simultaneously on lamps is 2 (e.g. at the moment 2).

In the second example, all lights are initially on. So the answer is 4.

### Response

```cpp
#include <bits/stdc++.h>
#pragma gcc optimize("unroll-loop")
#pragma gcc optimize("Ofast")
using namespace std;
const long long MOD = 1e9 + 7;
const long long N = 1e6 + 5;
long long gcd(long long n, long long m) {
  if (n == 0)
    return m;
  else
    return gcd(m % n, n);
}
long long ans[N], a[N], b[N];
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long n;
  cin >> n;
  string s;
  cin >> s;
  s = '#' + s;
  long long sum = 0;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i] >> b[i];
    long long light_on = 0;
    if (s[i] == '1') light_on = 1;
    for (long long j = 0; j <= 500; j++) {
      if (j < b[i]) {
        if (light_on == 1)
          ans[j] += 1;
        else
          ans[j] += 0;
      } else {
        if ((j - b[i]) % a[i] == 0) {
          ans[j] += light_on ^ 1;
          light_on ^= 1;
        } else {
          ans[j] += light_on;
        }
      }
    }
  }
  for (long long i = 0; i <= 500; i++) {
    sum = max(sum, ans[i]);
  }
  cout << sum << endl;
}

```