DZY loves colors, and he enjoys painting.

On a colorful day, DZY gets a colorful ribbon, which consists of n units (they are numbered from 1 to n from left to right). The color of the i-th unit of the ribbon is i at first. It is colorful enough, but we still consider that the colorfulness of each unit is 0 at first.

DZY loves painting, we know. He takes up a paintbrush with color x and uses it to draw a line on the ribbon. In such a case some contiguous units are painted. Imagine that the color of unit i currently is y. When it is painted by this paintbrush, the color of the unit becomes x, and the colorfulness of the unit increases by |x - y|.

DZY wants to perform m operations, each operation can be one of the following:

  1. Paint all the units with numbers between l and r (both inclusive) with color x. 
  2. Ask the sum of colorfulness of the units between l and r (both inclusive). 



Can you help DZY?

Input

The first line contains two space-separated integers n, m (1 ≤ n, m ≤ 105).

Each of the next m lines begins with a integer type (1 ≤ type ≤ 2), which represents the type of this operation.

If type = 1, there will be 3 more integers l, r, x (1 ≤ l ≤ r ≤ n; 1 ≤ x ≤ 108) in this line, describing an operation 1.

If type = 2, there will be 2 more integers l, r (1 ≤ l ≤ r ≤ n) in this line, describing an operation 2.

Output

For each operation 2, print a line containing the answer — sum of colorfulness.

Examples

Input

3 3
1 1 2 4
1 2 3 5
2 1 3


Output

8


Input

3 4
1 1 3 4
2 1 1
2 2 2
2 3 3


Output

3
2
1


Input

10 6
1 1 5 3
1 2 7 9
1 10 10 11
1 3 8 12
1 1 10 3
2 1 10


Output

129

Note

In the first sample, the color of each unit is initially [1, 2, 3], and the colorfulness is [0, 0, 0].

After the first operation, colors become [4, 4, 3], colorfulness become [3, 2, 0].

After the second operation, colors become [4, 5, 5], colorfulness become [3, 3, 2].

So the answer to the only operation of type 2 is 8.
#include <bits/stdc++.h>
using namespace std;
const int mod = 1000 * 1000 * 1000 + 7;
const int INF = 1e9 + 100;
const long long LINF = 1e18 + 100;
int n, m;
set<int> s;
int col[100100];
long long seg[4 * 100100];
long long lz[4 * 100100];
void pushdown(int v, int l, int r) {
  if (!lz[v]) return;
  int mid = (r + l) / 2;
  lz[2 * v] += lz[v];
  lz[2 * v + 1] += lz[v];
  seg[2 * v] += (mid - l + 1) * lz[v];
  seg[2 * v + 1] += (r - mid) * lz[v];
  lz[v] = 0;
}
void add(int L, int R, long long x, int v = 1, int l = 1, int r = n) {
  if (l > R || r < L) return;
  if (l >= L && r <= R) {
    seg[v] += (r - l + 1) * x;
    lz[v] += x;
    return;
  }
  int mid = (r + l) / 2;
  pushdown(v, l, r);
  add(L, R, x, 2 * v, l, mid);
  add(L, R, x, 2 * v + 1, mid + 1, r);
  seg[v] = seg[2 * v] + seg[2 * v + 1];
}
long long get(int L, int R, int v = 1, int l = 1, int r = n) {
  if (l > R || r < L) return 0;
  if (l >= L && r <= R) return seg[v];
  int mid = (r + l) / 2;
  pushdown(v, l, r);
  return get(L, R, 2 * v, l, mid) + get(L, R, 2 * v + 1, mid + 1, r);
}
int32_t main(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> m;
  for (int i = (1); i < (n + 1); i++) {
    s.insert(i);
    col[i] = i;
  }
  while (m--) {
    int op, l, r;
    cin >> op >> l >> r;
    if (op == 1) {
      int x;
      cin >> x;
      if (r < n && !s.count(r + 1)) {
        auto it = s.upper_bound(r);
        it--;
        col[r + 1] = col[*it];
        s.insert(r + 1);
      }
      auto it = s.upper_bound(l);
      it--;
      col[l] = col[*it];
      s.insert(l);
      it = s.lower_bound(l);
      while (it != s.end() && *it <= r) {
        auto it2 = it;
        it++;
        int hi = (it == s.end() ? n + 1 : *it);
        add(*it2, hi - 1, abs(col[*it2] - x));
        col[*it2] = x;
        if (*it2 > l) s.erase(it2);
      }
    } else {
      cout << get(l, r) << '\n';
    }
  }
  return 0;
}
