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>
using namespace std;
inline char nc() {
  static char buf[100000], *l = buf, *r = buf;
  return l == r && (r = (l = buf) + fread(buf, 1, 100000, stdin), l == r)
             ? EOF
             : *l++;
}
template <class T>
void read(T &x) {
  x = 0;
  int f = 1, ch = nc();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = nc();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 - '0' + ch;
    ch = nc();
  }
  x *= f;
}
const int inf = 1e9;
const int maxn = 1e5 + 5;
const int maxm = 1e6 + 5;
int n, m;
int dfc;
int anc[maxn];
int dep[maxn];
int dfn[maxn];
int rnk[maxn];
int siz[maxn];
int son[maxn];
int top[maxn];
int val[maxn];
int head[maxn];
struct edge {
  int to, nex, cost;
  edge(int to = 0, int nex = 0, int cost = 0) : to(to), nex(nex), cost(cost) {}
};
vector<edge> G;
inline void addedge(int u, int v, int w) {
  G.push_back(edge(v, head[u], w)), head[u] = G.size() - 1;
  G.push_back(edge(u, head[v], w)), head[v] = G.size() - 1;
}
struct data {
  int a, b, e;
  int id;
  inline bool operator<(const data &other) const { return e < other.e; }
} e[maxm];
inline int cmp(const data &a, const data &b) { return a.id < b.id; }
namespace us {
int fa[maxn];
void init(int n) {
  for (int i = 1; i <= n; ++i) fa[i] = i;
}
int find(int a) { return a == fa[a] ? a : fa[a] = find(fa[a]); }
bool merge(int a, int b) {
  a = find(a);
  b = find(b);
  if (a == b) return 0;
  fa[a] = b;
  return 1;
}
}  // namespace us
namespace seg {
int tag[maxn << 2];
int mxv[maxn << 2];
inline void pushup(int u) { mxv[u] = max(mxv[u << 1], mxv[u << 1 | 1]); }
inline void change(int u, int d) { tag[u] = min(tag[u], d); }
inline void pushdown(int u) {
  if (tag[u] != inf) {
    change(u << 1, tag[u]);
    change(u << 1 | 1, tag[u]);
    tag[u] = inf;
  }
}
void build(int u, int l, int r) {
  tag[u] = inf;
  if (l == r) {
    mxv[u] = val[rnk[l]];
    return;
  }
  int mid = (l + r) >> 1;
  build(u << 1, l, mid);
  build(u << 1 | 1, mid + 1, r);
  pushup(u);
}
void update(int u, int l, int r, int ql, int qr, int qv) {
  if (l == ql && r == qr) {
    change(u, qv);
    return;
  }
  int mid = (l + r) >> 1;
  pushdown(u);
  if (qr <= mid)
    update(u << 1, l, mid, ql, qr, qv);
  else if (ql > mid)
    update(u << 1 | 1, mid + 1, r, ql, qr, qv);
  else {
    update(u << 1, l, mid, ql, mid, qv);
    update(u << 1 | 1, mid + 1, r, mid + 1, qr, qv);
  }
}
int Qmin(int u, int l, int r, int qp) {
  if (l == r) {
    return tag[u];
  }
  int mid = (l + r) >> 1;
  pushdown(u);
  if (qp <= mid)
    return Qmin(u << 1, l, mid, qp);
  else
    return Qmin(u << 1 | 1, mid + 1, r, qp);
}
int Qmax(int u, int l, int r, int ql, int qr) {
  if (l == ql && r == qr) {
    return mxv[u];
  }
  int mid = (l + r) >> 1;
  if (qr <= mid)
    return Qmax(u << 1, l, mid, ql, qr);
  else if (ql > mid)
    return Qmax(u << 1 | 1, mid + 1, r, ql, qr);
  else {
    int L = Qmax(u << 1, l, mid, ql, mid);
    int R = Qmax(u << 1 | 1, mid + 1, r, mid + 1, qr);
    return max(L, R);
  }
}
}  // namespace seg
void update(int u, int v, int w) {
  while (top[u] != top[v]) {
    if (dep[top[u]] < dep[top[v]]) swap(u, v);
    seg ::update(1, 1, n, dfn[top[u]], dfn[u], w);
    u = anc[top[u]];
  }
  if (dep[u] < dep[v]) swap(u, v);
  if (u != v) seg ::update(1, 1, n, dfn[v] + 1, dfn[u], w);
}
int query(int u, int v) {
  int re = 0;
  while (top[u] != top[v]) {
    if (dep[top[u]] < dep[top[v]]) swap(u, v);
    re = max(re, seg ::Qmax(1, 1, n, dfn[top[u]], dfn[u]));
    u = anc[top[u]];
  }
  if (dep[u] < dep[v]) swap(u, v);
  if (u != v) re = max(re, seg ::Qmax(1, 1, n, dfn[v] + 1, dfn[u]));
  return re;
}
void dfs1(int u) {
  siz[u] = 1;
  for (int i = head[u]; ~i; i = G[i].nex) {
    int v = G[i].to;
    if (v == anc[u]) continue;
    anc[v] = u;
    val[v] = G[i].cost;
    dep[v] = dep[u] + 1;
    dfs1(v);
    siz[u] += siz[v];
    if (siz[v] >= siz[son[u]]) son[u] = v;
  }
}
void dfs2(int u, int chain) {
  top[u] = chain;
  dfn[u] = ++dfc;
  rnk[dfc] = u;
  if (son[u]) {
    dfs2(son[u], chain);
  }
  for (int i = head[u]; ~i; i = G[i].nex) {
    int v = G[i].to;
    if (v == son[u] || v == anc[u]) continue;
    dfs2(v, v);
  }
}
void kruskal() {
  us ::init(n);
  sort(e + 1, e + m + 1);
  memset(head, -1, sizeof(head));
  for (int i = 1; i <= m; ++i) {
    if (us ::merge(e[i].a, e[i].b)) {
      addedge(e[i].a, e[i].b, e[i].e);
    }
  }
}
int main() {
  read(n), read(m);
  for (int i = 1; i <= m; ++i) {
    read(e[i].a), read(e[i].b), read(e[i].e);
    e[i].id = i;
  }
  kruskal();
  dfs1(1);
  dfs2(1, 1);
  seg ::build(1, 1, n);
  for (int i = 1; i <= m; ++i) {
    if (anc[e[i].a] == e[i].b || anc[e[i].b] == e[i].a) continue;
    update(e[i].a, e[i].b, e[i].e);
  }
  sort(e + 1, e + m + 1, cmp);
  for (int i = 1; i <= m; ++i) {
    if (anc[e[i].a] == e[i].b || anc[e[i].b] == e[i].a) {
      printf("%d\n",
             seg ::Qmin(1, 1, n,
                        anc[e[i].a] == e[i].b ? dfn[e[i].a] : dfn[e[i].b]));
    } else {
      printf("%d\n", query(e[i].a, e[i].b));
    }
  }
  return 0;
}
