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 MAX = 2e5 + 9;
class LazySegmentTree {
 public:
  void reset(int sz, bool clr = false) {
    n = sz;
    treeMn.assign(n << 2, INT_MAX);
    treeMx.assign(n << 2, INT_MAX);
    lazyAssign.assign(n << 2, -1);
  }
  void minimize(int L, int R, int value) { minimize(1, n, 1, L, R, value); }
  int getMax(int L, int R) { return getMax(1, n, 1, L, R); }

 private:
  void push(int low, int high, int pos) {
    if (lazyAssign[pos] != -1) {
      treeMn[pos] = treeMx[pos] = lazyAssign[pos];
      if (low != high) {
        lazyAssign[pos << 1] = lazyAssign[pos << 1 | 1] = lazyAssign[pos];
      }
      lazyAssign[pos] = -1;
    }
  }
  void minimize(int low, int high, int pos, int L, int R, int value) {
    push(low, high, pos);
    if (L > high || R < low || treeMx[pos] <= value) {
      return;
    }
    if (L <= low && R >= high && treeMn[pos] >= value) {
      lazyAssign[pos] = value;
      push(low, high, pos);
      return;
    }
    int mid = (low + high) >> 1;
    minimize(low, mid, pos << 1, L, R, value);
    minimize(mid + 1, high, pos << 1 | 1, L, R, value);
    treeMn[pos] = min(treeMn[pos << 1], treeMn[pos << 1 | 1]);
    treeMx[pos] = max(treeMx[pos << 1], treeMx[pos << 1 | 1]);
  }
  int getMax(int low, int high, int pos, int L, int R) {
    push(low, high, pos);
    if (L > high || R < low) {
      return 0;
    }
    if (L <= low && R >= high) {
      return treeMx[pos];
    }
    int mid = (low + high) >> 1;
    return max(getMax(low, mid, pos << 1, L, R),
               getMax(mid + 1, high, pos << 1 | 1, L, R));
  }
  int n;
  vector<int> treeMx;
  vector<int> treeMn;
  vector<int> lazyAssign;
};
class Hld {
 public:
  void build(int n, const vector<vector<pair<int, int> > >& adj) {
    dfsCnt = 0;
    tree1.reset(n);
    tree2.reset(n);
    head.resize(n + 1);
    heavy.resize(n + 1);
    pos.resize(n + 1);
    subtreeSize.resize(n + 1);
    parent.resize(n + 1);
    depth.resize(n + 1);
    dfs(1, adj);
    decompose(1, 1, INT_MAX, adj);
  }
  void minimizePath(int x, int y, int value) {
    for (; head[x] != head[y]; y = parent[head[y]]) {
      if (depth[head[x]] > depth[head[y]]) {
        swap(x, y);
      }
      tree2.minimize(pos[head[y]], pos[y], value);
    }
    if (depth[x] > depth[y]) {
      swap(x, y);
    }
    if (x != y) {
      tree2.minimize(pos[x] + 1, pos[y], value);
    }
  }
  int getEdgeValue(int x, int y) {
    int position = max(pos[x], pos[y]);
    return tree2.getMax(position, position);
  }
  int getMaxOnPath(int x, int y) {
    int value = 0;
    for (; head[x] != head[y]; y = parent[head[y]]) {
      if (depth[head[x]] > depth[head[y]]) {
        swap(x, y);
      }
      value = max(value, tree1.getMax(pos[head[y]], pos[y]));
    }
    if (depth[x] > depth[y]) {
      swap(x, y);
    }
    if (x != y) {
      value = max(value, tree1.getMax(pos[x] + 1, pos[y]));
    }
    return value;
  }

 private:
  void dfs(int node, const vector<vector<pair<int, int> > >& adj) {
    int maxSubtreeSize = 0;
    subtreeSize[node] = 1;
    heavy[node].first = 0;
    for (const pair<int, int>& p : adj[node]) {
      int child = p.first;
      int cost = p.second;
      if (child != parent[node]) {
        parent[child] = node;
        depth[child] = depth[node] + 1;
        dfs(child, adj);
        subtreeSize[node] += subtreeSize[child];
        if (subtreeSize[child] > maxSubtreeSize) {
          maxSubtreeSize = subtreeSize[child];
          heavy[node] = {child, cost};
        }
      }
    }
  }
  void decompose(int node, int chainRoot, int edge,
                 const vector<vector<pair<int, int> > >& adj) {
    head[node] = chainRoot;
    pos[node] = ++dfsCnt;
    tree1.minimize(pos[node], pos[node], edge);
    if (heavy[node].first) {
      decompose(heavy[node].first, chainRoot, heavy[node].second, adj);
    }
    for (const pair<int, int>& p : adj[node]) {
      int child = p.first;
      int cost = p.second;
      if (child != parent[node] && child != heavy[node].first) {
        decompose(child, child, cost, adj);
      }
    }
  }
  int dfsCnt;
  vector<int> head;
  vector<pair<int, int> > heavy;
  vector<int> pos;
  vector<int> subtreeSize;
  vector<int> parent;
  vector<int> depth;
  LazySegmentTree tree1;
  LazySegmentTree tree2;
};
void Kruskal(int n, vector<vector<pair<int, int> > >& adj,
             vector<pair<pair<int, int>, pair<int, int> > >& edges,
             vector<bool>& isInMst) {
  vector<int> parent(n + 1);
  vector<int> r(n + 1);
  for (int i = 1; i <= n; ++i) {
    parent[i] = i;
  }
  function<int(int)> findSet = [&parent, &findSet](int node) -> int {
    if (parent[node] == node) {
      return node;
    } else {
      return parent[node] = findSet(parent[node]);
    }
  };
  auto mergeSets = [&findSet, &parent, &r](int x, int y) {
    int firstParent = findSet(x);
    int secondParent = findSet(y);
    assert(firstParent != secondParent);
    if (r[firstParent] > r[secondParent]) {
      swap(firstParent, secondParent);
    } else if (r[firstParent] == r[secondParent]) {
      ++r[secondParent];
    }
    parent[firstParent] = secondParent;
  };
  vector<pair<pair<int, int>, pair<int, int> > > tempEdges(edges);
  sort(tempEdges.begin(), tempEdges.end());
  for (int i = 0; i < tempEdges.size(); ++i) {
    int x = tempEdges[i].second.first;
    int y = tempEdges[i].second.second;
    int w = tempEdges[i].first.first;
    int idx = tempEdges[i].first.second;
    if (findSet(x) != findSet(y)) {
      adj[x].push_back({y, w});
      adj[y].push_back({x, w});
      isInMst[idx] = true;
      mergeSets(x, y);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m;
  cin >> n >> m;
  vector<pair<pair<int, int>, pair<int, int> > > edges(m);
  for (int i = 0; i < m; ++i) {
    int x, y, w;
    cin >> x >> y >> w;
    edges[i] = {{w, i}, {x, y}};
  }
  vector<vector<pair<int, int> > > adj(n + 1);
  vector<bool> isInMst(m);
  Kruskal(n, adj, edges, isInMst);
  Hld hld;
  hld.build(n, adj);
  vector<int> answer(m, INT_MAX);
  for (int i = 0; i < m; ++i) {
    if (!isInMst[i]) {
      int x = edges[i].second.first;
      int y = edges[i].second.second;
      int w = edges[i].first.first;
      answer[i] = hld.getMaxOnPath(x, y) - 1;
      hld.minimizePath(x, y, w - 1);
    }
  }
  for (int i = 0; i < m; ++i) {
    if (isInMst[i]) {
      int x = edges[i].second.first;
      int y = edges[i].second.second;
      answer[i] = hld.getEdgeValue(x, y);
      if (answer[i] == INT_MAX) {
        answer[i] = -1;
      }
    }
  }
  for (int i = 0; i < m; ++i) {
    cout << answer[i] << " ";
  }
  return 0;
}
