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 long long maxN = 1e5 + 7, MOD = 1e9 + 7, INF = 1e18 + 7, N = 5000 + 7,
                base = 707;
long long n, m, x, type, l, r, dp[maxN], data[4 * maxN], helper[4 * maxN], beg,
    fin;
set<int> q;
vector<int> vd, vc;
inline long long left(long long x) { return 2 * x + 1; }
inline long long right(long long x) { return 2 * x + 2; }
void up(long long node, long long st, long long en) {
  if (!helper[node] || en - st < 2) return;
  long long mid = (st + en) >> 1;
  helper[left(node)] += helper[node];
  helper[right(node)] += helper[node];
  data[left(node)] += (mid - st) * helper[node];
  data[right(node)] += (en - mid) * helper[node];
  helper[node] = 0;
}
void add(long long node, long long st, long long en, long long l, long long r,
         long long val) {
  if (en <= l || r <= st) return;
  if (st >= l && en <= r) {
    data[node] += (en - st) * val;
    helper[node] += val;
    return;
  }
  long long mid = (st + en) >> 1;
  up(node, st, en);
  add(left(node), st, mid, l, r, val);
  add(right(node), mid, en, l, r, val);
  data[node] = data[left(node)] + data[right(node)];
}
long long get(long long node, long long st, long long en, long long l,
              long long r) {
  if (en <= l || r <= st) return 0;
  if (st >= l && en <= r) return data[node];
  up(node, st, en);
  long long mid = (st + en) >> 1;
  return get(left(node), st, mid, l, r) + get(right(node), mid, en, l, r);
}
int main() {
  cin.tie(0), cout.tie(0), ios::sync_with_stdio(false);
  set<int>::iterator it, po;
  cin >> n >> m;
  for (int i = 0; i < n; i++) dp[i] = i + 1, q.insert(i);
  while (m--) {
    cin >> type >> l >> r;
    l--;
    if (type == 1) {
      cin >> x;
      r--;
      it = q.lower_bound(l);
      if (*it != l) it--;
      vd.clear();
      vc.clear();
      vd.push_back(l);
      if (r != n - 1) vd.push_back(r + 1);
      for (it; it != q.end(); it++) {
        beg = *it;
        po = q.end();
        po--;
        if (it != po) {
          po = it;
          po++;
          fin = *po;
          fin--;
        } else
          fin = n - 1;
        if (beg > r) break;
        if (beg <= r && beg >= l) vc.push_back(beg);
        if (beg >= l && fin <= r) {
          vc.push_back(beg);
          add(0, 0, n, beg, fin + 1, abs(x - dp[beg]));
        } else {
          if (beg < l && fin > r) {
            add(0, 0, n, l, r + 1, abs(x - dp[beg]));
            vd.push_back(r + 1);
            dp[r + 1] = dp[beg];
          } else {
            if (beg < l)
              add(0, 0, n, l, fin + 1, abs(x - dp[beg]));
            else {
              vd.push_back(r + 1);
              vc.push_back(beg);
              dp[r + 1] = dp[beg];
              add(0, 0, n, beg, r + 1, abs(x - dp[beg]));
            }
          }
        }
      }
      for (int i = 0; i < vc.size(); i++) q.erase(vc[i]);
      for (int i = 0; i < vd.size(); i++) q.insert(vd[i]);
      dp[l] = x;
    } else
      cout << get(0, 0, n, l, r) << "\n";
  }
  return cout << "\n", 0;
}
