#include <bits/stdc++.h>
using namespace std;
const long long int MOD = 1000000007;
const long long BIG = 1446803456761533460LL;
const int Big = 336860180;
const long long int INF = LONG_LONG_MAX;
const long long int adj4[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
const long long int adj8[8][2] = {{0, 1},   {0, -1}, {1, 0},  {-1, 0},
                                  {-1, -1}, {1, -1}, {-1, 1}, {1, 1}};
long long int clk_ar[10];
long long int gcd(long long int a, long long int b) {
  return b == 0 ? a : gcd(b, a % b);
}
long long int powMod(long long int a, long long int b) {
  long long int n = 1;
  long long int p = a;
  while (b > 0) {
    if (b % 2 == 1) {
      n *= p;
      n %= MOD;
    }
    p *= p;
    p %= MOD;
    b /= 2;
  }
  return n;
}
long long int modularInverse(long long int a) { return powMod(a, MOD - 2); }
stringstream sss;
const long long int maxn = 110;
const long long int T = 20000;
long long int state[maxn];
long long int ai[maxn];
long long int bi[maxn];
void MAIN() {
  long long int n;
  cin >> n;
  string s;
  cin >> s;
  for (long long int i = 0; i < (n); ++i) {
    state[i] = s[i] == '1';
  }
  for (long long int i = 0; i < (n); ++i) {
    cin >> ai[i] >> bi[i];
  }
  long long int mx = count(state, state + n, 1);
  for (long long int t = 0; t < (T); ++t) {
    for (long long int i = 0; i < (n); ++i) {
      if (t >= bi[i] && (t - bi[i]) % ai[i] == 0) {
        state[i] ^= 1;
      }
    }
    mx = max(mx, ((long long int)count(state, state + n, 1)));
  }
  cout << (mx) << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(10);
  sss << R"(
6
011100
5 3
5 5
2 4
3 5
4 2
1 5
    )";
  MAIN();
  return 0;
}

### Prompt

Please create a solution in cpp to the following problem:

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>
using namespace std;
const long long int MOD = 1000000007;
const long long BIG = 1446803456761533460LL;
const int Big = 336860180;
const long long int INF = LONG_LONG_MAX;
const long long int adj4[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
const long long int adj8[8][2] = {{0, 1},   {0, -1}, {1, 0},  {-1, 0},
                                  {-1, -1}, {1, -1}, {-1, 1}, {1, 1}};
long long int clk_ar[10];
long long int gcd(long long int a, long long int b) {
  return b == 0 ? a : gcd(b, a % b);
}
long long int powMod(long long int a, long long int b) {
  long long int n = 1;
  long long int p = a;
  while (b > 0) {
    if (b % 2 == 1) {
      n *= p;
      n %= MOD;
    }
    p *= p;
    p %= MOD;
    b /= 2;
  }
  return n;
}
long long int modularInverse(long long int a) { return powMod(a, MOD - 2); }
stringstream sss;
const long long int maxn = 110;
const long long int T = 20000;
long long int state[maxn];
long long int ai[maxn];
long long int bi[maxn];
void MAIN() {
  long long int n;
  cin >> n;
  string s;
  cin >> s;
  for (long long int i = 0; i < (n); ++i) {
    state[i] = s[i] == '1';
  }
  for (long long int i = 0; i < (n); ++i) {
    cin >> ai[i] >> bi[i];
  }
  long long int mx = count(state, state + n, 1);
  for (long long int t = 0; t < (T); ++t) {
    for (long long int i = 0; i < (n); ++i) {
      if (t >= bi[i] && (t - bi[i]) % ai[i] == 0) {
        state[i] ^= 1;
      }
    }
    mx = max(mx, ((long long int)count(state, state + n, 1)));
  }
  cout << (mx) << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(10);
  sss << R"(
6
011100
5 3
5 5
2 4
3 5
4 2
1 5
    )";
  MAIN();
  return 0;
}

```