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;
struct ed {
  int u, v, cost, idx;
  ed() {}
  ed(int uu, int vv, int cc, int idx) : u(uu), v(vv), cost(cc), idx(idx) {}
  bool operator<(const ed& a) const { return cost < a.cost; }
};
struct edge {
  int next, cost;
  edge() {}
  edge(int a, int c) : next(a), cost(c) {}
};
struct node {
  int height;
  int par[23];
  int chainNum;
  int heavyNode;
  int segNum;
  int cost[23];
  int childcnt;
};
int n, m;
node info[200001];
int chainHead[200001];
vector<edge> adj[200001];
bool visited[200001];
void dfs(int cur, int height) {
  int cnt = 1;
  int maxV = 0;
  int maxnode = 0;
  for (auto& a : adj[cur]) {
    if (!visited[a.next]) {
      visited[a.next] = true;
      info[a.next].height = height + 1;
      info[a.next].par[0] = cur;
      info[a.next].cost[0] = a.cost;
      dfs(a.next, height + 1);
      if (info[a.next].childcnt > maxV) {
        maxV = info[a.next].childcnt;
        maxnode = a.next;
      }
      cnt += info[a.next].childcnt;
    }
  }
  info[cur].heavyNode = maxnode;
  info[cur].childcnt = cnt;
  return;
}
void processLca(int n) {
  for (int i = 1; i < 23; i++)
    for (int j = 1; j <= n; j++) {
      info[j].par[i] = info[info[j].par[i - 1]].par[i - 1];
      info[j].cost[i] =
          max(info[info[j].par[i - 1]].cost[i - 1], info[j].cost[i - 1]);
    }
}
vector<int> seg;
vector<int> lazy;
void lazy_update(int idx, int s, int e) {
  if (lazy[idx] == (1987654321)) return;
  if (lazy[idx] < seg[idx]) {
    seg[idx] = lazy[idx];
    lazy[idx] = (1987654321);
    if (s != e) {
      lazy[idx * 2] = min(lazy[idx * 2], seg[idx]);
      lazy[idx * 2 + 1] = min(lazy[idx * 2 + 1], seg[idx]);
    }
  }
  return;
}
void seg_update(int idx, int s, int e, int l, int r, int val) {
  lazy_update(idx, s, e);
  if (e < l || s > r) return;
  if (l <= s && e <= r) {
    seg[idx] = min(seg[idx], val);
    if (s != e) {
      lazy[idx * 2] = min(lazy[idx * 2], val);
      lazy[idx * 2 + 1] = min(lazy[idx * 2 + 1], val);
    }
    return;
  }
  int mid = (s + e) / 2;
  seg_update(idx * 2, s, mid, l, r, val);
  seg_update(idx * 2 + 1, mid + 1, e, l, r, val);
  return;
}
int seg_min(int idx, int s, int e, int l, int r) {
  lazy_update(idx, s, e);
  if (e < l || s > r) return (1987654321);
  if (l <= s && e <= r) return seg[idx];
  int mid = (s + e) >> 1;
  int a = seg_min(idx * 2, s, mid, l, r);
  int b = seg_min(idx * 2 + 1, mid + 1, e, l, r);
  return min(a, b);
}
int chainCnt = 1;
int segcnt = 1;
int NodeNum[200001];
void HLD(int cur) {
  NodeNum[segcnt] = cur;
  info[cur].segNum = segcnt++;
  info[cur].chainNum = chainCnt;
  if (info[cur].heavyNode != 0) {
    visited[info[cur].heavyNode] = true;
    HLD(info[cur].heavyNode);
  }
  for (auto& a : adj[cur]) {
    if (!visited[a.next]) {
      visited[a.next] = true;
      chainCnt++;
      chainHead[chainCnt] = a.next;
      HLD(a.next);
    }
  }
  return;
}
int getLca(int a, int b) {
  int ha = info[a].height;
  int hb = info[b].height;
  if (ha != hb) {
    int cha;
    if (ha > hb) {
      swap(hb, ha);
      swap(a, b);
    }
    cha = hb - ha;
    for (int i = 22; i >= 0; i--) {
      if (cha & (1 << i)) {
        b = info[b].par[i];
      }
    }
  }
  if (a == b)
    return a;
  else {
    for (int i = 22; i >= 0; i--) {
      if (info[a].par[i] != info[b].par[i])
        a = info[a].par[i], b = info[b].par[i];
    }
  }
  return info[a].par[0];
}
int getMax(int lca, int b) {
  int hlc = info[lca].height;
  int hb = info[b].height;
  int cha = hb - hlc;
  int maxV = 0;
  for (int i = 23; i >= 0; i--) {
    if (cha & (1 << i)) {
      maxV = max(maxV, info[b].cost[i]);
      b = info[b].par[i];
    }
  }
  return maxV;
}
void update(int lca, int b, int v) {
  while (info[lca].chainNum != info[b].chainNum) {
    seg_update(1, 1, n, info[chainHead[info[b].chainNum]].segNum,
               info[b].segNum, v);
    b = info[chainHead[info[b].chainNum]].par[0];
  }
  if (b == lca) return;
  seg_update(1, 1, n, info[lca].segNum + 1, info[b].segNum, v);
}
int getMin(int a) { return seg_min(1, 1, n, info[a].segNum, info[a].segNum); }
ed edL[200001];
int par[200001];
int rr[200001];
bool inMst[200001];
int find(int a) {
  if (par[a] == a) return a;
  return par[a] = find(par[a]);
}
void merge(int a, int b) {
  int pa = find(a);
  int pb = find(b);
  if (rr[pa] > rr[pb]) swap(pa, pb);
  par[pa] = pb;
  if (rr[pa] == rr[pb]) rr[pb]++;
}
void MakeKruscal() {
  for (int i = 0; i < m; i++) {
    if (find(edL[i].u) == find(edL[i].v))
      inMst[edL[i].idx] = false;
    else {
      inMst[edL[i].idx] = true;
      merge(edL[i].u, edL[i].v);
      adj[edL[i].u].push_back(edge(edL[i].v, edL[i].cost));
      adj[edL[i].v].push_back(edge(edL[i].u, edL[i].cost));
    }
  }
}
int ans[200001];
int main(void) {
  scanf("%d %d", &n, &m);
  for (int i = 0, a, b, c; i < m; i++) {
    scanf("%d %d %d", &a, &b, &c);
    edL[i] = ed(a, b, c, i);
  }
  sort(edL, edL + m);
  for (int i = 1; i <= n; i++) par[i] = i;
  MakeKruscal();
  visited[1] = true;
  info[1].height = 1;
  dfs(1, 1);
  processLca(n);
  memset(visited, false, sizeof(visited));
  visited[1] = true;
  chainHead[chainCnt] = 1;
  HLD(1);
  seg.resize(533333, (1987654321));
  lazy.resize(533333, (1987654321));
  for (int i = 0; i < m; i++) {
    if (!inMst[edL[i].idx]) {
      int lc = getLca(edL[i].u, edL[i].v);
      ans[edL[i].idx] = max(getMax(lc, edL[i].u), getMax(lc, edL[i].v));
      update(lc, edL[i].u, edL[i].cost);
      update(lc, edL[i].v, edL[i].cost);
    }
  }
  for (int i = 0; i < m; i++) {
    if (inMst[edL[i].idx]) {
      if (info[edL[i].u].par[0] == edL[i].v)
        ans[edL[i].idx] = getMin(edL[i].u);
      else
        ans[edL[i].idx] = getMin(edL[i].v);
    }
  }
  for (int i = 0; i < m; i++) {
    if (ans[i] == (1987654321))
      printf("-1 ");
    else
      printf("%d ", ans[i] - 1);
  }
  return 0;
}
