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;
int n, m;
long long sum[411111], add[411111];
set<pair<pair<int, int>, int> > W;
set<pair<pair<int, int>, int> >::iterator it;
void mod(int pos, int l, int r, int ll, int rr, long long v) {
  ll = max(ll, l);
  rr = min(rr, r);
  if (ll > rr) return;
  if (l == ll && r == rr) {
    sum[pos] += (r - l + 1) * v;
    add[pos] += v;
    return;
  }
  if (add[pos] != 0) {
    mod(pos + pos, l, (l + r) / 2, l, r, add[pos]);
    mod(pos + pos + 1, (l + r) / 2 + 1, r, l, r, add[pos]);
    add[pos] = 0;
  }
  mod(pos + pos, l, (l + r) / 2, ll, rr, v);
  mod(pos + pos + 1, (l + r) / 2 + 1, r, ll, rr, v);
  sum[pos] = sum[pos + pos] + sum[pos + pos + 1];
}
long long fin(int pos, int l, int r, int ll, int rr) {
  ll = max(ll, l);
  rr = min(rr, r);
  if (ll > rr) return 0;
  if (l == ll && r == rr) return sum[pos];
  if (add[pos] != 0) {
    mod(pos + pos, l, (l + r) / 2, l, r, add[pos]);
    mod(pos + pos + 1, (l + r) / 2 + 1, r, l, r, add[pos]);
    add[pos] = 0;
  }
  long long q1 = fin(pos + pos, l, (l + r) / 2, ll, rr);
  long long q2 = fin(pos + pos + 1, (l + r) / 2 + 1, r, ll, rr);
  return q1 + q2;
}
void di(int x) {
  it = W.lower_bound(make_pair(make_pair(x + 1, x), 0));
  if (it == W.begin()) return;
  it--;
  int ll = (*it).first.first;
  int rr = (*it).first.second;
  int col = (*it).second;
  if (x >= ll && x < rr) {
    W.erase(it);
    W.insert(make_pair(make_pair(ll, x), col));
    W.insert(make_pair(make_pair(x + 1, rr), col));
  }
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < n; i++) W.insert(make_pair(make_pair(i, i), i));
  while (m--) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      int l, r, x;
      scanf("%d%d%d", &l, &r, &x);
      l--;
      r--;
      x--;
      di(l - 1);
      di(r);
      for (;;) {
        it = W.lower_bound(make_pair(make_pair(l, l), 0));
        if (it == W.end() || (*it).first.first > r) break;
        int dif = abs(x - (*it).second);
        mod(1, 0, n - 1, (*it).first.first, (*it).first.second, dif);
        W.erase(it);
      }
      W.insert(make_pair(make_pair(l, r), x));
    } else {
      int l, r;
      scanf("%d%d", &l, &r);
      l--;
      r--;
      printf("%I64d\n", fin(1, 0, n - 1, l, r));
    }
  }
  return 0;
}
