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;
const int maxn = 400010, inf = 1 << 30;
int N, M, father[maxn], nxt[maxn], toit[maxn], len[maxn], side[maxn], id[maxn],
    seq[maxn], D[maxn], cnt;
int DFN, dfn[maxn], top[maxn], size[maxn], con[maxn], ans[maxn],
    tree[maxn << 2], cover[maxn << 2];
bool exist[maxn];
inline int find(int a) {
  if (father[a] != a) father[a] = find(father[a]);
  return father[a];
}
inline void add(int a, int b, int c, int d) {
  nxt[++cnt] = side[a];
  side[a] = cnt;
  toit[cnt] = b;
  len[cnt] = c;
  id[cnt] = d;
}
inline void ins(int a, int b, int c, int d) {
  add(a, b, c, d);
  add(b, a, c, d);
}
inline int gi() {
  char ch;
  int ret = 0, f = 1;
  do ch = getchar();
  while (!(ch >= '0' && ch <= '9') && ch != '-');
  if (ch == '-') f = -1, ch = getchar();
  do ret = ret * 10 + ch - '0', ch = getchar();
  while (ch >= '0' && ch <= '9');
  return ret * f;
}
struct Edge {
  int u, v, c, id;
  inline void read(int i) {
    id = i;
    u = gi(), v = gi(), c = gi();
  }
  inline void bind() {
    int a = find(u), b = find(v);
    if (a != b) {
      father[a] = b;
      exist[id] = true;
      ins(u, v, c, id);
    }
  }
  friend inline bool operator<(const Edge &a, const Edge &b) {
    return a.c < b.c;
  }
} edge[maxn];
inline void Div(int now, int Top) {
  dfn[now] = ++DFN;
  seq[DFN] = now;
  top[now] = Top;
  int heavy = 0;
  for (int i = side[now]; i; i = nxt[i])
    if (toit[i] != father[now] && size[toit[i]] > size[heavy]) heavy = toit[i];
  if (!heavy) return;
  Div(heavy, Top);
  for (int i = side[now]; i; i = nxt[i])
    if (toit[i] != father[now] && toit[i] != heavy) Div(toit[i], toit[i]);
}
inline void dfs(int now) {
  size[now] = 1;
  for (int i = side[now]; i; i = nxt[i])
    if (toit[i] != father[now]) {
      father[toit[i]] = now, con[toit[i]] = i;
      D[toit[i]] = D[now] + 1, dfs(toit[i]);
      size[now] += size[toit[i]];
    }
}
inline void mst() {
  sort(edge + 1, edge + M + 1);
  cnt = 1;
  for (int i = 1; i <= N; ++i) father[i] = i;
  for (int i = 1; i <= M; ++i) edge[i].bind();
  father[1] = 0;
  dfs(1);
  Div(1, 1);
}
inline int ask(int now, int l, int r, int ql, int qr) {
  if (l == ql && r == qr) return tree[now];
  int mid = (l + r) >> 1;
  if (qr <= mid)
    return ask(now << 1, l, mid, ql, qr);
  else if (ql > mid)
    return ask(now << 1 | 1, mid + 1, r, ql, qr);
  else
    return max(ask(now << 1, l, mid, ql, mid),
               ask(now << 1 | 1, mid + 1, r, mid + 1, qr));
}
inline void modify(int now, int l, int r, int ql, int qr, int c) {
  if (l == ql && r == qr) {
    cover[now] = min(cover[now], c);
    return;
  }
  int mid = (l + r) >> 1;
  if (qr <= mid)
    modify(now << 1, l, mid, ql, qr, c);
  else if (ql > mid)
    modify(now << 1 | 1, mid + 1, r, ql, qr, c);
  else {
    modify(now << 1, l, mid, ql, mid, c);
    modify(now << 1 | 1, mid + 1, r, mid + 1, qr, c);
  }
}
inline int Query(int a, int b, int c) {
  int ret = -inf;
  while (top[a] != top[b]) {
    if (D[top[a]] < D[top[b]]) swap(a, b);
    ret = max(ret, ask(1, 1, N, dfn[top[a]], dfn[a]));
    modify(1, 1, N, dfn[top[a]], dfn[a], c);
    a = father[top[a]];
  }
  if (D[a] < D[b]) swap(a, b);
  if (a != b) {
    ret = max(ret, ask(1, 1, N, dfn[b] + 1, dfn[a]));
    modify(1, 1, N, dfn[b] + 1, dfn[a], c);
  }
  return ret;
}
inline void build(int now, int l, int r) {
  cover[now] = inf;
  if (l == r) {
    tree[now] = len[con[seq[l]]];
    return;
  }
  int mid = (l + r) >> 1;
  build(now << 1, l, mid);
  build(now << 1 | 1, mid + 1, r);
  tree[now] = max(tree[now << 1], tree[now << 1 | 1]);
}
inline int query(int now, int l, int r, int pos) {
  if (l == r) return cover[now];
  int mid = (l + r) >> 1;
  if (pos <= mid)
    return min(cover[now], query(now << 1, l, mid, pos));
  else
    return min(cover[now], query(now << 1 | 1, mid + 1, r, pos));
}
inline void work() {
  build(1, 1, N);
  for (int i = 1; i <= M; ++i)
    if (!exist[edge[i].id])
      ans[edge[i].id] = Query(edge[i].u, edge[i].v, edge[i].c);
  for (int i = 2; i <= N; ++i) ans[id[con[i]]] = query(1, 1, N, dfn[i]);
  for (int i = 1; i <= M; ++i) {
    if (ans[i] == inf)
      ans[i] = -1;
    else
      --ans[i];
    printf("%d ", ans[i]);
  }
}
int main() {
  N = gi();
  M = gi();
  for (int i = 1; i <= M; ++i) edge[i].read(i), ans[i] = inf;
  mst();
  work();
  return 0;
}
