5. 155 birds are perched on a circle $C$. If $\overparen{P_{i} P_{j}} \leqslant 10^{\circ}$, we say $P_{i}$ and $P_{j}$ form a mutually visible pair of birds. Find the minimum number of mutually visible pairs of birds (assuming multiple birds can occupy the same position).

 untranslated text:
(assuming a position can simultaneously have several birds).
5. Let $A, B$ be two points on a circle, bird $P_{i}$ is at point $A$, and bird $P$ is at point $B$, $\overparen{A B} \leqslant 10^{\circ}$. Let $k$ be the number of birds that can be seen from $B$ but not from $A$, and $l$ be the number of birds that can be seen from $A$ but not from $B$, with $k \geqslant l$. If all the birds at $B$ are moved to $A$, then for each bird at $B$, the number of visible pairs decreases by $k$ and increases by $l$, so the number of mutually visible pairs decreases by $k-l$. After such adjustments, the number of bird positions decreases by 1, and the number of mutually visible pairs does not increase. Repeating this adjustment several times, we can make it so that birds are only mutually visible when they are at the same position. At this point, the number of bird positions is at most 35, so the problem is reduced to finding $\min \left\{\sum_{i=1}^{35} C_{x_{i}}^{2} ; \sum_{i=1}^{35} x_{i}=155, x_{i} \geqslant 0\right\}$. Assume $x_{1}=\min x_{i}, x_{2}=\max x_{i}$, if $x_{2}-x_{1} \geqslant 2$, then let $x_{2}^{\prime}=x_{2}-1, x_{1}^{\prime}=x_{1}+1$, and keep other $x_{i}$ unchanged, then the sum $\sum_{i=1}^{35} x_{i}$ remains unchanged, $\mathrm{C}_{x_{1}}^{2}+\mathrm{C}_{x_{2}}^{2}-\mathrm{C}_{x_{1}^{\prime}}^{2}-\mathrm{C}_{x_{2}^{\prime}}^{2}=\frac{x_{1}\left(x_{1}-1\right)}{2}+\frac{x_{2}\left(x_{2}-1\right)}{2}-\frac{x_{1}\left(x_{1}+1\right)}{2}-\frac{\left(x_{2}-1\right)\left(x_{2}-2\right)}{2}=-x_{1}+x_{2}-1 \geqslant 1$, so after the adjustment, $\sum_{i=1}^{35} \mathrm{C}_{x_{i}}^{2}$ decreases. To minimize $\sum_{i=1}^{35} \mathrm{C}_{x_{i}}^{2}$, we should have $x_{2}-x_{1} \leqslant 1$. Since $4<\frac{155}{35}<5$, $155=20 \times 4+15 \times 5$, so $\sum_{i=1}^{35} \mathrm{C}_{x_{i}}^{2}=20 \times \mathrm{C}_{4}^{2}+15 \times \mathrm{C}_{5}^{2}=270$.