#include <bits/stdc++.h>
using namespace std;
mt19937 eng;
int n, m;
string s;
basic_string<int> e[100005];
int sm[100005], mp[3];
vector<array<int, 3>> sol;
bool st[100005], vis[100005];
int dopuna(int x) {
  x %= 3;
  if (x < 0) x += 3;
  return 3 - x;
}
void dfs(int x, int par) {
  st[x] = vis[x] = 1;
  for (int y : e[x]) {
    if (y == par) continue;
    if (!vis[y]) {
      dfs(y, x);
      int t = dopuna(sm[y] - mp[s[y] - 'X']);
      sol.push_back({x, y, t});
      sm[x] += t;
      sm[y] += t;
    } else if (st[y]) {
      sm[x] += 3;
      sm[y] += 3;
      sol.push_back({x, y, 3});
    }
  }
  st[x] = 0;
}
bool probaj() {
  iota(mp, mp + 3, 0);
  for (int i = 0; i < n; i++) shuffle(e[i].begin(), e[i].end(), eng);
  shuffle(mp, mp + 3, eng);
  fill(sm, sm + n, 0);
  fill(vis, vis + n, 0);
  int x = uniform_int_distribution<int>(0, n - 1)(eng);
  sol = {};
  dfs(x, x);
  for (int i = 0; i < n; i++)
    for (int j : e[i])
      if (sm[i] == sm[j]) return false;
  for (auto a : sol) cout << a[0] << ' ' << a[1] << ' ' << a[2] << '\n';
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cerr.tie(nullptr);
  cin >> n >> m >> s;
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    e[u] += v;
    e[v] += u;
  }
  while (!probaj())
    ;
}

### Prompt

Your challenge is to write a Cpp solution to the following problem:

Three planets X, Y and Z within the Alpha planetary system are inhabited with an advanced civilization. The spaceports of these planets are connected by interplanetary space shuttles. The flight scheduler should decide between 1, 2 and 3 return flights for every existing space shuttle connection. Since the residents of Alpha are strong opponents of the symmetry, there is a strict rule that any two of the spaceports connected by a shuttle must have a different number of flights. 

For every pair of connected spaceports, your goal is to propose a number 1, 2 or 3 for each shuttle flight, so that for every two connected spaceports the overall number of flights differs. 

You may assume that:

1) Every planet has at least one spaceport 

2) There exist only shuttle flights between spaceports of different planets 

3) For every two spaceports there is a series of shuttle flights enabling traveling between them

4) Spaceports are not connected by more than one shuttle

Input

The first row of the input is the integer number N (3 ≤ N ≤ 100 000), representing overall number of spaceports. The second row is the integer number M (2 ≤ M ≤ 100 000) representing number of shuttle flight connections. 

Third row contains N characters from the set \\{X, Y, Z\}. Letter on I^{th} position indicates on which planet is situated spaceport I. For example, "XYYXZZ" indicates that the spaceports 0 and 3 are located at planet X, spaceports 1 and 2 are located at Y, and spaceports 4 and 5 are at Z. 

Starting from the fourth row, every row contains two integer numbers separated by a whitespace. These numbers are natural numbers smaller than N and indicate the numbers of the spaceports that are connected. For example, "12\ 15" indicates that there is a shuttle flight between spaceports 12 and 15. 

Output

The same representation of shuttle flights in separate rows as in the input, but also containing a third number from the set \{1, 2, 3\} standing for the number of shuttle flights between these spaceports.

Example

Input


10
15
XXXXYYYZZZ
0 4
0 5
0 6
4 1
4 8
1 7
1 9
7 2
7 5
5 3
6 2
6 9
8 2
8 3
9 3


Output


0 4 2
0 5 2
0 6 2
4 1 1
4 8 1
1 7 2
1 9 3
7 2 2
7 5 1
5 3 1
6 2 1
6 9 1
8 2 3
8 3 1
9 3 1

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
mt19937 eng;
int n, m;
string s;
basic_string<int> e[100005];
int sm[100005], mp[3];
vector<array<int, 3>> sol;
bool st[100005], vis[100005];
int dopuna(int x) {
  x %= 3;
  if (x < 0) x += 3;
  return 3 - x;
}
void dfs(int x, int par) {
  st[x] = vis[x] = 1;
  for (int y : e[x]) {
    if (y == par) continue;
    if (!vis[y]) {
      dfs(y, x);
      int t = dopuna(sm[y] - mp[s[y] - 'X']);
      sol.push_back({x, y, t});
      sm[x] += t;
      sm[y] += t;
    } else if (st[y]) {
      sm[x] += 3;
      sm[y] += 3;
      sol.push_back({x, y, 3});
    }
  }
  st[x] = 0;
}
bool probaj() {
  iota(mp, mp + 3, 0);
  for (int i = 0; i < n; i++) shuffle(e[i].begin(), e[i].end(), eng);
  shuffle(mp, mp + 3, eng);
  fill(sm, sm + n, 0);
  fill(vis, vis + n, 0);
  int x = uniform_int_distribution<int>(0, n - 1)(eng);
  sol = {};
  dfs(x, x);
  for (int i = 0; i < n; i++)
    for (int j : e[i])
      if (sm[i] == sm[j]) return false;
  for (auto a : sol) cout << a[0] << ' ' << a[1] << ' ' << a[2] << '\n';
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cerr.tie(nullptr);
  cin >> n >> m >> s;
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    e[u] += v;
    e[v] += u;
  }
  while (!probaj())
    ;
}

```