#include <bits/stdc++.h>
using namespace std;
pair<long long, long long> operator+(const pair<long long, long long> &a,
                                     const pair<long long, long long> &b) {
  return make_pair(a.first + b.first, a.second + b.second);
}
pair<long long, long long> operator-(const pair<long long, long long> &a,
                                     const pair<long long, long long> &b) {
  return make_pair(a.first - b.first, a.second - b.second);
}
pair<long long, long long> operator*(const pair<long long, long long> &a,
                                     const long long &b) {
  return make_pair(a.first * b, a.second * b);
}
ostream &operator<<(ostream &os, const pair<long long, long long> &p) {
  os << p.first << ' ' << p.second;
  return os;
}
const long double PI = acos(-1);
const long long INF = (long long)(1e18) + 9;
const long double EPS = 1e-15;
const long double SEPS = 1e-6;
const long long NN = 80050;
const long long N = 1024 * 128;
const long long K = 4096;
const long long L = 26;
const unsigned long long INTF = (long long)INT_MAX * INT_MAX;
const long long LET = 26;
const long long MUL = 1000;
const long long F = 4;
long long n;
string s;
vector<long long> a, b;
vector<bool> c;
signed main() {
  double start = clock();
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  cout << fixed << setprecision(13);
  cerr << setprecision(10);
  srand(14313);
  {}
  cin >> n;
  cin >> s;
  for (long long i = 0; i < n; i++) c.push_back(s[i] == '1');
  a.resize(n);
  b.resize(n);
  for (long long i = 0; i < n; i++) cin >> a[i] >> b[i];
  long long ans = 0;
  for (long long i = 0; i < N; i++) {
    long long cnt = 0;
    for (long long j = 0; j < n; j++) {
      if (i >= b[j] && i % a[j] == b[j] % a[j]) c[j] = c[j] ^ 1;
      if (c[j]) cnt++;
    }
    ans = max(ans, cnt);
  }
  cout << ans;
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
pair<long long, long long> operator+(const pair<long long, long long> &a,
                                     const pair<long long, long long> &b) {
  return make_pair(a.first + b.first, a.second + b.second);
}
pair<long long, long long> operator-(const pair<long long, long long> &a,
                                     const pair<long long, long long> &b) {
  return make_pair(a.first - b.first, a.second - b.second);
}
pair<long long, long long> operator*(const pair<long long, long long> &a,
                                     const long long &b) {
  return make_pair(a.first * b, a.second * b);
}
ostream &operator<<(ostream &os, const pair<long long, long long> &p) {
  os << p.first << ' ' << p.second;
  return os;
}
const long double PI = acos(-1);
const long long INF = (long long)(1e18) + 9;
const long double EPS = 1e-15;
const long double SEPS = 1e-6;
const long long NN = 80050;
const long long N = 1024 * 128;
const long long K = 4096;
const long long L = 26;
const unsigned long long INTF = (long long)INT_MAX * INT_MAX;
const long long LET = 26;
const long long MUL = 1000;
const long long F = 4;
long long n;
string s;
vector<long long> a, b;
vector<bool> c;
signed main() {
  double start = clock();
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  cout << fixed << setprecision(13);
  cerr << setprecision(10);
  srand(14313);
  {}
  cin >> n;
  cin >> s;
  for (long long i = 0; i < n; i++) c.push_back(s[i] == '1');
  a.resize(n);
  b.resize(n);
  for (long long i = 0; i < n; i++) cin >> a[i] >> b[i];
  long long ans = 0;
  for (long long i = 0; i < N; i++) {
    long long cnt = 0;
    for (long long j = 0; j < n; j++) {
      if (i >= b[j] && i % a[j] == b[j] % a[j]) c[j] = c[j] ^ 1;
      if (c[j]) cnt++;
    }
    ans = max(ans, cnt);
  }
  cout << ans;
  return 0;
}

```