With your help, Heidi has prepared a plan of trap placement and defence. Yet suddenly, the Doctor popped out of the TARDIS and told her that he had spied on the Daleks' preparations, and there is more of them than ever. Desperate times require desperate measures, so Heidi is going to risk meeting with the Daleks and she will consider placing a trap along any Corridor.

This means she needs your help again in calculating E_{max}(c) – the largest e ≤ 10^9 such that if we changed the energy requirement of c to e, then the Daleks might use c in their invasion – but this time for all Time Corridors.

Input

First line: number n of destinations, number m of corridors (2 ≤ n ≤ 10^5, n - 1 ≤ m ≤ 10^6). The next m lines: destinations a, b and energy e (1 ≤ a, b ≤ n, a ≠ b, 0 ≤ e ≤ 10^9).

No pair \\{a, b\} will repeat. The graph is guaranteed to be connected. It is not guaranteed that all energy requirements e are distinct, or that the minimum spanning tree is unique.

Output

Output m lines, each containing one integer: E_{max}(c_i) for the i-th Corridor c_i from the input.

Example

Input

3 3
1 2 8
2 3 3
3 1 4


Output

4
8
8
#include <bits/stdc++.h>
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx")
#pragma GCC optimize 03
#pragma GCC optimize("unroll-loops")
using namespace std;
int gcd(int a, int b) { return (a == 0) ? b : gcd(b % a, a); }
int cnt1(int x) {
  int cnt = 0;
  while (x > 0) x &= x - 1, cnt++;
  return cnt;
}
int fact(int n) { return n == 0 ? 1LL : n * fact(n - 1); }
int fact(int n, int p) { return n == 0 ? 1LL : n * fact(n - 1) % p; }
int cnk(int n, int k) {
  return n == 1 || k == 1 ? 1 : cnk(n - 1, k) + cnk(n - 1, k - 1);
}
int pow(int a, int n, int p) {
  if (n == 0) return 1;
  long long t = pow(a, n >> 1, p);
  return t * t % p * (n & 1 ? a : 1) % p;
}
int back(int x, int p) { return pow(x, p - 2, p); }
int cnk(int n, int k, int p) {
  return fact(n, p) * back(fact(k, p), p) % p * back(fact(n - k, p), p) % p;
}
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
template <class T>
istream &operator>>(istream &stream, vector<T> &v) {
  for (int i = 0; i < (int)v.size(); i++) stream >> v[i];
  return stream;
}
template <class T>
ostream &operator<<(ostream &stream, vector<T> v) {
  for (int i = 0; i < (int)v.size(); i++) stream << v[i] << ' ';
  stream << '\n';
  return stream;
}
const int N = 1e+5 + 1;
int p[N];
int find_p(int v) {
  if (p[v] == v)
    return v;
  else
    return p[v] = find_p(p[v]);
}
bool union_set(int a, int b) {
  a = find_p(a);
  b = find_p(b);
  if (a == b) return 0;
  if (rnd() & 1)
    p[a] = b;
  else
    p[b] = a;
  return 1;
}
vector<pair<pair<int, int>, pair<int, int>>> e;
vector<pair<int, int>> gr[N];
const int M = 1e+6;
int ans[M];
bool in_mst[M];
set<pair<int, int>> s[N];
int ns[N];
int tin[N], tout[N], timer;
void dfs(int v, int p = -1, int laste = -1) {
  tin[v] = timer++;
  int mx = v, umx = v;
  for (auto x : gr[v]) {
    int u = x.first, ne = x.second;
    if (u != p) {
      if (in_mst[ne]) {
        dfs(u, v, e[ne].first.second);
        if (s[ns[u]].size() > s[mx].size()) mx = ns[u], umx = u;
      } else
        s[v].insert({e[ne].first.first, u});
    }
  }
  tout[v] = timer;
  if (s[v].size() > s[mx].size()) mx = v, umx = v;
  ns[v] = mx;
  for (auto x : gr[v]) {
    int u = x.first, ne = x.second;
    if (u != p && u != umx)
      if (in_mst[ne]) {
        for (auto t : s[ns[u]]) s[mx].insert(t);
        s[ns[u]].clear();
      }
  }
  if (mx != v) {
    for (auto t : s[v]) s[mx].insert(t);
    s[v].clear();
  }
  while (s[mx].size() && tin[v] <= tin[s[mx].begin()->second] &&
         tin[s[mx].begin()->second] < tout[v])
    s[mx].erase(s[mx].begin());
  if (laste != -1)
    if (s[mx].size() == 0)
      ans[laste] = 1e+9;
    else
      ans[laste] = s[mx].begin()->first;
}
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    a--, b--;
    e.push_back({{c, i}, {a, b}});
  }
  vector<int> l(m), r(m);
  for (int i = 0; i < m; i++) r[i] = m;
  sort(e.begin(), e.end());
  for (int t = 0; t < 20; t++) {
    vector<pair<int, int>> q(m);
    for (int i = 0; i < m; i++) q[i] = {(l[i] + r[i]) >> 1, i};
    sort(q.begin(), q.end());
    for (int i = 0; i < n; i++) p[i] = i;
    int i, j = 0;
    for (i = 0; i < m; i++) {
      if (union_set(e[i].second.first, e[i].second.second)) in_mst[i] = 1;
      while (j < m && q[j].first == i) {
        if (find_p(e[q[j].second].second.first) ==
            find_p(e[q[j].second].second.second))
          r[q[j].second] = i;
        else
          l[q[j].second] = i;
        j++;
      }
    }
  }
  for (int i = 0; i < m; i++)
    if (!in_mst[i]) ans[e[i].first.second] = e[l[i] + 1].first.first;
  for (int i = 0; i < m; i++) {
    gr[e[i].second.first].push_back({e[i].second.second, i});
    gr[e[i].second.second].push_back({e[i].second.first, i});
  }
  dfs(0);
  for (int i = 0; i < m; i++) cout << ans[i] << '\n';
}
