// 解法2
#include <bits/stdc++.h>
using namespace std;

static const int INF = 2e9;

// Segment-Tree
struct SegmentTree {
  int n;
  vector<int> dat, lazy;
  
  SegmentTree(int _n) {
    init(_n);
  }
  
  void init(int _n) {
    n = 1;
    while (n < _n) n *= 2;
    dat.assign(2 * n - 1, INF);
    lazy.assign(2 * n - 1, INF);
  }
  
  // k番目の値を更新(RUQ)
  void eval(int k) {
    if (lazy[k] == INF) return;
    if (k < n - 1) {
      lazy[2 * k + 1] = lazy[k];
      lazy[2 * k + 2] = lazy[k];
    }
    dat[k] = lazy[k];
    lazy[k] = INF;
  }
  
  // 区間[a, b)の値をxに変更(RUQ)
  void update(int a, int b, int x, int k = 0, int l = 0, int r = -1) {
    if (r == -1) r = n;
    eval(k);
    if (a <= l && r <= b) {
      lazy[k] = x;
      eval(k);
    }
    else if (a < r && l < b) {
      update(a, b, x, 2 * k + 1, l, (l + r) / 2);
      update(a, b, x, 2 * k + 2, (l + r) / 2, r);
      dat[k] = min(dat[2 * k + 1], dat[2 * k + 2]);
    }
  }
  
  // [a, b)の値. ノードkの範囲は[l, r)
  int query(int a, int b, int k = 0, int l = 0, int r = -1) {
    if (r == -1) r = n;
    if (r <= a || b <= l) return INF;
    eval(k); // RUQ
    if (a <= l && r <= b) return dat[k];
    else {
      int vl = query(a, b, 2 * k + 1, l, (l + r) / 2);
      int vr = query(a, b, 2 * k + 2, (l + r) / 2, r);
      return min(vl, vr);
    }
  }
};

int main() {
  int n, q;
  cin >> n >> q;
  
  vector<vector<int>> query(q, vector<int>(3));
  for (int i = 0; i < q; i++) {
    cin >> query.at(i).at(1) >> query.at(i).at(2) >> query.at(i).at(0);
    query.at(i).at(1)--;
    query.at(i).at(2)--;
  }
  
  sort(query.begin(), query.end());
  
  SegmentTree st(n);
  for (int i = 0; i < q; i++) {
    st.update(query.at(i).at(1), query.at(i).at(2) + 1, query.at(i).at(0));
  }
  
  bool ok = true;
  for (int i = 0; i < q; i++) {
    if (st.query(query.at(i).at(1), query.at(i).at(2) + 1) != query.at(i).at(0)) {
      ok = false;
    }
  }
  
  if (ok) {
    for (int i = 0; i < n; i++) {
      if (i) cout << " ";
      cout << min<int>(st.query(i, i + 1), 1e9);
    }
    cout << '\n';
  }
  else {
    cout << -1 << '\n';
  }
}