You are given a connected weighted graph with n vertices and m edges. The graph doesn't contain loops nor multiple edges. Consider some edge with id i. Let's determine for this edge the maximum integer weight we can give to it so that it is contained in all minimum spanning trees of the graph if we don't change the other weights.

You are to determine this maximum weight described above for each edge. You should calculate the answer for each edge independently, it means there can't be two edges with changed weights at the same time.

Input

The first line contains two integers n and m (2 ≤ n ≤ 2·105, n - 1 ≤ m ≤ 2·105), where n and m are the number of vertices and the number of edges in the graph, respectively.

Each of the next m lines contains three integers u, v and c (1 ≤ v, u ≤ n, v ≠ u, 1 ≤ c ≤ 109) meaning that there is an edge between vertices u and v with weight c. 

Output

Print the answer for each edge in the order the edges are given in the input. If an edge is contained in every minimum spanning tree with any weight, print -1 as the answer.

Examples

Input

4 4
1 2 2
2 3 2
3 4 2
4 1 3


Output

2 2 2 1 

Input

4 3
1 2 2
2 3 2
3 4 2


Output

-1 -1 -1 
#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long num = 0, neg = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') neg = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    num = (num << 3) + (num << 1) + c - '0';
    c = getchar();
  }
  return num * neg;
}
const int INF = 0x3f3f3f3f;
struct node {
  int x, y, c, id;
} s[200010];
bool cmp1(node x, node y) { return x.c < y.c; }
bool book[200010];
int n, m, dep[200010], fa[200010], size[200010], son[200010], p[200010];
int id[200010], tmp[200010], tot, top[200010], ans[200010];
int bel[200010], to[400010], head[200010], nxt[400010], val[400010], cnt;
int a[200010 << 2], b[200010 << 2], tag[200010 << 2];
inline int find(int x) { return bel[x] == x ? x : bel[x] = find(bel[x]); }
inline void insert(int x, int y, int z) {
  to[++cnt] = y, nxt[cnt] = head[x], head[x] = cnt, val[cnt] = z;
}
inline void kruskal() {
  sort(s + 1, s + m + 1, cmp1);
  for (int i = 1; i <= n; i++) bel[i] = i;
  for (int i = 1; i <= m; i++) {
    int u = find(s[i].x), v = find(s[i].y);
    if (u != v)
      book[i] = 1, bel[u] = v, insert(s[i].x, s[i].y, s[i].c),
      insert(s[i].y, s[i].x, s[i].c);
  }
  return;
}
inline void dfs1(int u, int father) {
  dep[u] = dep[father] + 1, fa[u] = father, size[u] = 1;
  int maxson = 0;
  for (int i = head[u]; i; i = nxt[i]) {
    int v = to[i];
    if (v == father) continue;
    p[v] = val[i];
    dfs1(v, u);
    size[u] += size[v];
    if (size[v] > size[maxson]) maxson = v;
  }
  son[u] = maxson;
  return;
}
inline void dfs2(int u, int topf) {
  id[u] = ++tot, tmp[tot] = p[u], top[u] = topf;
  if (!son[u]) return;
  dfs2(son[u], topf);
  for (int i = head[u]; i; i = nxt[i]) {
    int v = to[i];
    if (v == fa[u] || v == son[u]) continue;
    dfs2(v, v);
  }
  return;
}
inline void push_up(int x) {
  a[x] = max(a[x << 1], a[x << 1 | 1]), b[x] = min(b[x << 1], b[x << 1 | 1]);
}
inline void build(int l, int r, int x) {
  if (l == r) {
    a[x] = tmp[l], b[x] = INF;
    return;
  }
  int mid = l + r >> 1;
  build(l, mid, x << 1), build(mid + 1, r, x << 1 | 1);
  push_up(x);
}
inline void push_down(int x) {
  if (tag[x]) {
    b[x << 1] = b[x << 1 | 1] = tag[x << 1] = tag[x << 1 | 1] = tag[x];
    tag[x] = 0;
  }
  return;
}
inline void update(int ul, int ur, int l, int r, int x, int k) {
  if (ul <= l && ur >= r) {
    b[x] = tag[x] = k;
    return;
  }
  int mid = l + r >> 1;
  push_down(x);
  if (ul <= mid) update(ul, ur, l, mid, x << 1, k);
  if (ur > mid) update(ul, ur, mid + 1, r, x << 1 | 1, k);
  return;
}
inline int querya(int ql, int qr, int l, int r, int x) {
  if (ql <= l && qr >= r) return a[x];
  int mid = l + r >> 1, res = 0;
  push_down(x);
  if (ql <= mid) res = max(res, querya(ql, qr, l, mid, x << 1));
  if (qr > mid) res = max(res, querya(ql, qr, mid + 1, r, x << 1 | 1));
  return res;
}
inline int queryb(int q, int l, int r, int x) {
  if (l == r && q == l) return b[x];
  int mid = l + r >> 1;
  push_down(x);
  if (q <= mid)
    return queryb(q, l, mid, x << 1);
  else
    return queryb(q, mid + 1, r, x << 1 | 1);
}
inline void updrange(int x, int y, int k) {
  while (top[x] != top[y]) {
    if (dep[top[x]] < dep[top[y]]) swap(x, y);
    update(id[top[x]], id[x], 1, n, 1, k);
    x = fa[top[x]];
  }
  if (dep[x] > dep[y]) swap(x, y);
  if (x == y) return;
  update(id[x] + 1, id[y], 1, n, 1, k);
}
inline int qrange(int x, int y) {
  int res = 0;
  while (top[x] != top[y]) {
    if (dep[top[x]] < dep[top[y]]) swap(x, y);
    res = max(res, querya(id[top[x]], id[x], 1, n, 1));
    x = fa[top[x]];
  }
  if (dep[x] > dep[y]) swap(x, y);
  if (x == y) return res;
  res = max(res, querya(id[x] + 1, id[y], 1, n, 1));
  return res;
}
int main() {
  n = read(), m = read();
  for (int i = 1; i <= m; i++)
    s[i].x = read(), s[i].y = read(), s[i].c = read(), s[i].id = i;
  kruskal();
  dfs1(1, 1), dfs2(1, 1), build(1, n, 1);
  for (int i = m; i >= 1; i--) {
    if (!book[i]) {
      updrange(s[i].x, s[i].y, s[i].c);
      ans[s[i].id] = qrange(s[i].x, s[i].y) - 1;
    } else {
      int u = (dep[s[i].x] > dep[s[i].y] ? s[i].x : s[i].y);
      ans[s[i].id] = queryb(id[u], 1, n, 1) - 1;
    }
  }
  for (int i = 1; i <= m; i++) {
    if (ans[i] == INF - 1)
      printf("-1 ");
    else
      printf("%d ", ans[i]);
  }
  return 0;
}
